C1777tk

GeneralizedPartitionsandNewIdeasOnNumberTheoryandSmarandacheSequencesEditor'sNoteThisbookaroseoutofacollectionofpaperswrittenbyAmarnathMurthy.
ThepapersdealwithmathematicalideasderivedfromtheworkofFlorentinSmarandache,amanwhoseemstohavenoendofideas.
MostofthepaperswerepublishedinSmarandacheNotionsJournalandtherewasagreatdealofoverlap.
Myintentintransformingthepapersintoacoherentbookwastoremovetheduplications,organizethematerialbasedontopicandcleanupsomeofthemostobviouserrors.
However,Imadenoattempttoverifyeverystatement,sothemathematicalworkisalmostexclusivelythatofMurthy.
IwouldalsoliketothankTylerBrogla,whocreatedtheimagethatappearsonthefrontcover.
CharlesAshbacherSmarandacheRepeatableReciprocalPartitionofUnity{2,3,10,15}1/2+1/3+1/10+1/15=1AmarnathMurthy/CharlesAshbacher1AMARNATHMURTHYS.
E.
(E&T)WELLLOGGINGSERVICESOILANDNATURALGASCORPORATIONLTDCHANDKHEDAAHMEDABADGUJARAT-380005INDIACHARLESASHBACHERMOUNTMERCYCOLLEGE1330ELMHURSTDRIVENECEDARRAPIDS,IOWA42402USAGENERALIZEDPARTITIONSANDSOMENEWIDEASONNUMBERTHEORYANDSMARANDACHESEQUENCESHexisPhoenix20052Thisbookcanbeorderedinapaperboundreprintfrom:BooksonDemandProQuestInformation&Learning(UniversityofMicrofilmInternational)300N.
ZeebRoadP.
O.
Box1346,AnnArborMI48106-1346,USATel.
:1-800-521-0600(CustomerService)http://wwwlib.
umi.
com/bod/search/basicPeerReviewers:1)Eng.
MarianPopescu,Str.
Spaniei,Bl.
O8,Sc.
1,Ap.
14,Craiova,Jud.
Dolj,Romania.
2)Dr.
SukantoBhattacharya,DepartmentofBusinessAdministration,AlaskaPacificUniversity,U.
S.
A.
3)Dr.
M.
Khoshnevisan,SchoolofAccountingandFinance,GriffithUniversity,GoldCoast,Queensland9726,Australia.
Copyright2005byHexis.
,AmarnathMurthyandCharlesAshbacherManybookscanbedownloadedfromthefollowingE-LibraryofScience:http://www.
gallup.
unm.
edu/~smarandache/eBooks-otherformats.
htmFrontcoverimageTylerBroglaandCharlesAshbacherISBN:1-931233-34-9StandardAddressNumber:297-5092PrintedintheUnitedStatesofAmerica3Editor'sNoteThisbookaroseoutofacollectionofpaperswrittenbyAmarnathMurthy.
ThepapersdealwithmathematicalideasderivedfromtheworkofFlorentinSmarandache,amanwhoseemstohavenoendofideas.
MostofthepaperswerepublishedinSmarandacheNotionsJournalandtherewasagreatdealofoverlap.
Myintentintransformingthepapersintoacoherentbookwastoremovetheduplications,organizethematerialbasedontopicandcleanupsomeofthemostobviouserrors.
However,Imadenoattempttoverifyeverystatement,sothemathematicalworkisalmostexclusivelythatofMurthy.
IwouldalsoliketothankTylerBrogla,whocreatedtheimagethatappearsonthefrontcover.
CharlesAshbacher4TableofContentsEditor'sNote3TableofContents4Chapter1SmarandachePartitionFunctions……………….
111.
1SmarandachePartitionSets,SequencesandFunctions……………….
111.
2AProgramtoDeterminetheNumberofSmarandacheDistinctReciprocalPartitionsofUnityofaGivenLength……………….
161.
3ANoteOnMaohuaLe'sProofofMurthy'sConjectureOnReciprocalPartitionTheory……………….
181.
4GeneralizationofPartitionFunctions,IntroductionoftheSmarandachePartitionFunction……………….
191.
5OpenProblemsandConjecturesOntheFactor/ReciprocalPartitionTheory……………….
281.
6AGeneralResultontheSmarandacheStarFunction……………….
301.
7MoreResultsandApplicationsoftheGeneralizedSmarandacheStarFunction……………….
401.
8PropertiesoftheSmarandacheStarTriangle471.
9SmarandacheFactor505PartitionsOfATypicalCanonicalForm1.
10Length/ExtentofSmarandacheFactorPartitions551.
11MoreIdeasOnSmarandacheFactorPartitions…………………581.
12ANoteOntheSmarandacheDivisorSequences…………………611.
13AnAlgorithmforListingtheSmarandacheFactorPartitions…………………621.
14AProgramForDeterminingtheNumberofSFPs…………………641.
15ChapterReferences…………………70Chapter2SmarandacheSequences…………………722.
1OntheLargestBaluNumberandSomeSFPEquations…………………722.
2SmarandachePascalDerivedSequences…………………752.
3DepascalizationofSmarandachePascalDerivedSequencesandBackwardExtendedFibonacciSequence…………………812.
4ProofthetheDepascalizationTheorem…………………832.
5SmarandacheFriendlyNumbersandAFewMoreSequences…………………862.
6SomeNewSmarandacheSequences,FunctionsandPartitions…………………8962.
7SmarandacheReverseAutoCorrelatedSequencesandSomeFibonacciDerivedSmarandacheSequences…………………942.
8SmarandacheStar(Stirling)DerivedSequences…………………982.
9SmarandacheStrictlyStaircaseSequence…………………1002.
10TheSumoftheReciprocalsoftheSmarandacheMultiplicativeSequence…………………1012.
11DecompositionoftheDivisorsofANaturalNumberIntoPairwiseCo-PrimeSets…………………1032.
12OntheDivisorsoftheSmarandacheUnarySequence…………………1062.
13SmarandacheDualSymmetricFunctionsandCorrespondingNumbersoftheTypeofStirlingNumbersoftheFirstKind…………………1082.
14OntheInfinitudeoftheSmarandacheAdditiveSquareSequence…………………1102.
15OntheInfinitudeoftheSmarandacheMultiplicativeSquareSequence…………………1112.
16AnotherClassificationoftheOceanof…………………1137SmarandacheSequences2.
17PouringaFewDropsintheOceanofSmarandacheSequencesandSeries…………………1142.
18SmarandachePythagorasAdditiveSquareSequence1182.
19TheNumberofElementstheSmarandacheMultiplicativeSquareSequenceandtheSmarandacheAdditiveSquareSequenceHaveinCommon1202.
20SmarandachePatternedPerfectCubeSequences…………………1212.
21TheSmarandacheAdditiveCubeSequenceisInfinite…………………1222.
22MoreExamplesandResultsOntheInfinitudeofCertainSmarandacheSequences…………………1232.
23SmarandacheSymmetric(Palindromic)PerfectPowerSequences…………………1242.
24SomePropositionsOntheSmarandachen2nSequence…………………1262.
25TheSmarandacheFermatAdditiveCubeSequence…………………1282.
26TheSmarandachenn2SequenceContainsNoPerfect…………………1308Squares2.
27PrimesintheSmarandachennmSequence…………………1322.
28SomeIdeasOntheSmarandachenknSequence1332.
29SomeNotionsOnLeastCommonMultiples1342.
30AnApplicationoftheSmarandacheLCMSequenceandtheLargestNumberDivisiblebyAlltheIntegersNotExceedingItsrthRoot1372.
31TheNumberofPrimesIntheSmarandacheMultipleSequence1382.
32MoreOntheSmarandacheSquareandHigherPowerBases1392.
33SmarandacheFourthandHigherPatterned/AdditivePerfectPowerSequences1402.
34TheSmarandacheMultiplicativeCubicSequenceandMoreIdeasonDigitSums1422.
35SmarandachePrimeGeneratorSequence1432.
36ChapterReferences145Chapter3MiscellaneousTopics1473.
1ExploringSomeNewIdeasOnSmarandacheTypeSets,FunctionsandSequences14793.
2FabricatingPerfectSquaresWithaGivenValidDigitSum1553.
3FabricatingPerfectCubesWithaGivenValidDigitSum1573.
4SmarandachePerfectPowersWithGivenValidDigitSum1613.
5NumbersThatAreaMultipleoftheProductofTheirDigitsandRelatedIdeas1643.
6TheLargestandSmallestmthPowerWhoseDigitSum/ProductIsIt'smthRoot1663.
7AConjectureond(N),theDivisorFunctionItselfAsADivisorWithRequiredJustification1683.
8SmarandacheFitorialandSupplementaryFitorialFunctions1713.
9SomeMoreConjecturesOnPrimesandDivisors1743.
10SmarandacheReciprocalFunctionandAnElementaryInequality1763.
11SmarandacheMaximumReciprocalRepresentationFunction1783.
12SmarandacheDeterminantSequence179103.
13ExpansionofxninSmarandacheTermsofPermutations1823.
14MiscellaneousResultsandTheoremsonSmarandacheTermsandFactorPartitions1873.
15Smarandache-Murthy'sFiguresofPeriodicSymmetryofRotationSpecifictoanAngle…………………1923.
16SmarandacheRouteSequences1983.
17SmarandacheGeometricalPartitionsandSequences2003.
18SmarandacheLuckyMethodsinAlgebra,TrigonometryandCalculus2073.
19ChapterReferences210Index21211Chapter1SmarandachePartitionFunctionsSection1SmarandachePartitionSets,SequencesandFunctionsUnitfractionsarefractionswherethenumeratoris1andthedenominatorisanaturalnumber.
Ourfirstpointofinterestisindeterminingallsetsofunitfractionsofacertainsizewherethesumoftheelementsinthesetis1.
Definition:Forn>0,theSmarandacheRepeatableReciprocalpartitionofunityforn(SRRPS(n))isthesetofallsetsofnnaturalnumberssuchthatthesumofthereciprocalsis1.
Moreformally,nSRRPS(n)={x|x=(a1,a2,an)whereΣ1/ar=1}.
r=1fRP(n)=orderofthesetSRRPS(n).
Forexample,SRRPS(1)={(1)},fRP(1)=1.
SRRPS(2)={(2,2)},fRP(2)=1.
(1/2+1/2=1).
SRRPS(3)={(3,3,3),(2,3,6),(2,4,4)},fRP(3)=3.
SRRPS(4)={(4,4,4,4),(2,4,6,12),(2,3,7,42),(2,4,5,20),(2,6,6,6),(2,4,8,8),(2,3,12,12),(4,4,3,6),(3,3,6,6),(2,3,10,15),(2,3,9,18)},fRP(4)=14.
Definition:TheSmarandacheRepeatableReciprocalPartitionofUnitySequenceisthesequenceofnumbersSRRPS(1),SRRPS(2),SRRPS(3),SRRPS(4),SRRPS(5),.
.
.
Definition:Forn>0,theSmarandacheDistinctReciprocalPartitionofUnitySet(SDRPS(n))isSRRPS(n)wheretheelementsofeachsetofsizenmustbeunique.
Moreformally,nSRRPS(n)={x|x=(a1,a2,an)whereΣ1/ar=1andai=aji=j}.
r=1fDP(n)=orderofSDRPS(n).
Forexample:SRRPS(1)={(1)},fRP(1)=1.
SRRPS(2)fRP(2)=0.
12SRRPS(3)={(2,3,6)},fRP(3)=1.
SRRPS(4)={(2,4,6,12),(2,3,7,42),(2,4,5,20),(2,3,10,15),(2,3,9,18)},fRP(4)=5.
Definition:TheSmarandacheDistinctReciprocalpartitionofunitysequenceisthesequenceofnumbersfDP(n).
Theorem:n-1fDP(n)≥∑fDP(k)+(n2–5n+8)/2,n>3.
k=3Proof:Theinequalitywillbeestablishedintwosteps.
PropositionAForeveryn,thereexistsasetofndistinctnaturalnumbers,thesumofwhosereciprocalsis1.
ProofofpropositionA:Theproofisbyinductiononn.
Basisstep:n=1,1/1=1.
Inductivestep:Assumethatthepropositionistrueforr.
Thenthereisasetofdistinctnaturalnumbersa13.
k=3Remark:Itshouldbepossibletocomeupwithastrongerresult,asIbelievethatthereshouldbemoretermsontheright.
Thereasonforthisbeliefwillbeclearfromthefollowingtheorem.
Theorem:LetmbeamemberofanelementofSRRPS(n),saym=ak,from(a1,a2,an)andbydefinitionn∑1/ak=1.
k=1Themcontributes[(d(m)+1)/2]elementstoSRRPS(n+1),where[]representstheintegervalueandd(m)thenumberofdivisorsofm.
Proof:Foreachdivisordofm,thereisacorrespondingdivisorm/d=d'.
Case-I:misnotaperfectsquare.
Thend(m)isevenandthereared(m)/2pairsofthetype(d,d')suchthatdd'=m.
15Considerthefollowingidentity1/(p*q)=1/(p(p+q))+1/(q(p+q))foreachdivisorpair(d,d')ofmwehavethefollowingbreakup1/(d*d')=1/(d(d+d'))+1/(d'(d+d')).
Hence,thecontributionofmtoSRRPS(n+1)isd(m)/2.
Asd(m)isevend(m)/2=[(d(m)+1)/2]aswell.
Case-IImisaperfectsquare.
Inthiscased(m)isoddandthereisadivisorpaird=d'=m1/2.
ThiswillcontributeonetoSRRPS(n+1).
Theremaining(d(m)–1)/2pairsofdistinctdivisorswouldeachcontributeone,makingthetotalcontribution((d(m)–1)/2).
Therefore,thetotalnumberinthiscasewouldbe(d(m)–1)/2+1=(d(m)+1)/2=[(d(m)+1)/2].
Hencemcontributes[(d(m)+1)/2]elementstoSRRPS(n+1)andtheproofiscomplete.
Remarks:1)ThetotalcontributiontoSRRPS(n+1)byanyelementofSRRPS(n)is∑[(d(ak)+1)/2]whereeachakisconsideredonlyonceirrespectiveofitsrepeatedoccurrence.
2)ForSDRPS(n+1),thecontributionbyanyelementofSDRPS(n)isgivenby∑[d(ak)/2]becausethedivisorpaird=d'=akdoesnotcontribute.
Hence,thetotalcontributionofSDRP(n)togenerateSDRPS(n+1)isthesummationoveralltheelementsofSDRPS(n).
n∑[∑(d(ak)/2)].
fDP(n)k=1Generalization:Itispossibletogeneralizetheseresultsbyconsideringthefollowingidentity161111=++pqrpq(p+q+r)qr(p+q+r)rp(p+q+r)whichalsosuggests1rrr=∑[(Пbt)(∑bs)]-1b1b2.
.
.
brk=1t=1,t≠ks=1Itiseasytoestablishthisidentitybysumminguptheelementsontherightside.
Fromthisformula,thecontributionoftheelementsofSDRPS(n)toSDRPS(n+r)canbeevaluatedifanswerstothefollowingopenproblemscanbefound.
Openproblems:(1)Inhowmanywayscananumberbeexpressedastheproductof3ofitsdivisors(2)Ingeneralinhowmanywayscananumberbeexpressedastheproductofrofits'divisors(3)InhowmanywayscananumberbeexpressedastheproductofitsdivisorsAnyattempttofindanswerstotheabovequestionsleadstotheneedforthegeneralizationofthetheoryofthepartitionfunction.
Section2AProgramtoDeterminetheNumberofSmarandacheDistinctReciprocalPartitionsofUnityofaGivenLengthTheprevioussectionintroducedtheSmarandachedistinctreciprocalpartitionofunityanddemonstratedsomeproperties.
Inthissection,acomputerprogramwrittenintheClanguagewillbepresented.
/*ThisisaprogramforfindingnumberofdistinctreciprocalpartitionsofunityofagivenlengthwrittenbyKSuresh,Softwareexpert,IKOS,NOIDA,INDIA.
*/#include#includeunsignedlongTOTAL;FILE*f;longdoublearray[100];17unsignedlongcount=0;voidtry(longdoubleprod,longdoublesum,unsignedlongpos){if(pos==TOTAL-1){//lastelement.
.
longdoublediff=prod-sum;if(diff==0)return;array[pos]=floorl(prod/diff);if(array[pos]>array[pos-1]&&array[pos]*diff==prod){fprintf(f,"(%ld)%ld",++count,(unsignedlong)array[0]);inti;for(i=1;i=new_prod)continue;try(new_prod,new_sum,pos+1);}return;}18main(){printf("Enternoofelements");scanf("%ld",&TOTAL);charfname[256];sprintf(fname,"rec%ld.
out",TOTAL);f=fopen(fname,"w");fprintf(f,"Noofelements=%ld.
\n",TOTAL);try(1,0,0);fflush(f);fclose(f);printf("Total%ldsolutionsfound.
\n",count);return0;}Usingthisprogram,thefollowingtableofdatawasaccumulated.
LengthNumberofDistinctReciprocalPrimes11203146572623207245765Section3ANoteOnMaohuaLe'sProofOfMurthy'sConjectureOnReciprocalPartitionTheoryIn[4],MaohuaLeattemptedtoprovetheconjecturethatthereareinfinitelymanydisjointsetsofpositiveintegersthesumofwhosereciprocalsisequaltoone.
Hemisunderstoodtheconjecture,perhapsduetoinadequatewording.
Whatheactuallyprovedwasthepropositionthatforeverynthereexistsasetofndistinctnaturalnumbersthesumofwhosereciprocalsisone.
Iwouldliketoclarifyandrestatetheconjectureusingthefollowingexample.
LetA={a1,a2,ar}andB={b1,b2,bs}betwosetssuchthat19rs∑1/ak=1=∑1/bkk=1k=1withA∩B=.
TheconjectureisthatthereareinfinitelymanydisjointsetsofthetypeAorB.
Example:A={2,3,7,42},B={4,5,6,8,9,12,20,72}.
Section4GeneralizationofPartitionFunction,IntroductionoftheSmarandacheFactorPartitionThepartitionfunctionP(n)isdefinedasthenumberofwaysthatapositiveintegercanbeexpressedasthesumofpositiveintegers.
Twopartitionsarenotconsidereddifferentiftheydifferonlyintheorderoftheirsummands.
ManyresultsconcerningthepartitionfunctionwerediscoveredusinganalyticfunctionsbyEuler,Jacobi,Hardy,Ramanujanandothers.
Otherpropertiesofthefunctioninvolvingcongruencesarealsoknown.
Intheprevioussections,theconceptoftheSmarandacheReciprocalPartitionsofunitywasintroduced.
Oneoftheproblemsconsideredwasthenumberofwaysinwhichanumbercanbeexpressedastheproductofitsdivisors.
Inthissection,wewillexaminesomegeneralizationsoftheconceptofpartitioninganumber.
Definition:Letα1,α2,αrbeasetofnaturalnumbersandp1,p2,prasetofarbitraryprimes.
TheSmarandacheFactorPartition(SFP)of(α1,α2,αr),F(α1,α2,αr)isdefinedasthenumberofwaysinwhichthenumberN=p1α1p2α2.
.
.
prαrcanbeexpressedastheproductofits'divisors.
Example:Withthesetofprimes,(2,3),F(1,2)=4,asN=21*32=18(1)N=18(2)N=2*9(3)N=3*6(4)N=2*3*3.
ItisaconsequenceofthedefinitionofSFPandfactorsthatF(α1,α2)=F(α2,α1)andingeneraltheorderoftheαiin(α1,α2,αr)isimmaterial.
Also,theprimesp1,p2,.
.
.
,prcanbechosenarbitrarily.
20Theorem:F(m)=P(m),whereP(m)isthenumberofadditionpartitionsofm.
Proof:Letpbeanyprime,N=pmandm=x1+x2xnbeanadditionpartitionofm.
Then,N=(px1)(px2pxn)isaSFPofN,i.
e.
eachpartitionofmcontributesoneSFP.
Also,letoneoftheSFPofNbeN=(N1)(N2)Nk).
EachNihastobeoftheformNi=pai.
LetN1=pa1,N2=pa2pak.
ThenN=(pa1)(pa2pak)=p(a1+a2+…+ak)=>m=a1+a2akwhichgivesapartitionofm.
Obviously,eachSFPofNgivesoneuniquepartitionofm.
Therefore,sinceeachapproachyieldsoneSFP,F(m)=P(m).
αTheorem:F(α,1)=ΣP(k)k=0Proof:LetN=p1αp2,wherep1,p2arearbitrarilychosenprimes.
Case1:WritingN=(p2)p1α,keepingp2asaseparateentity,(oneofthefactorsinthefactorpartitionofN),bytheprevioustheoremwouldyieldP(α)Smarandachefactorpartitions.
Case2:WritingN=(p1p2)p1α-1keeping(p1p2)asaseparateentity(oneofthefactorsofSFPofN)wouldyieldP(α-1)SFPs.
.
.
.
Caser:Ingeneral,writingN=(p1rp2)pα-randkeeping(p1rp2)asaseparateentitywouldyieldP(α-r)SFPs.
ContributionstowardsF(N)ineachofthecasesaremutuallydisjointasp1rp2isuniqueforagivenr,whichrangesfrom0toα,whichisexhaustive.
Therefore,21αF(α,1)=ΣP(α-r)r=0Letα–r=k,r=0=>k=α,r=α=>k=0.
0F(α,1)=ΣP(k)k=ααF(α,1)=ΣP(k)k=0whichcompletestheproof.
Examples:I.
F(3)=P(3)=3,Letp=2,N=23=8.
(1)N=8,(2)N=4*2,(3)N=2*2*2.
4II.
F(4,1)=ΣP(k)=P(0)+P(1)+P(2)+P(3)+P(4)k=0=1+1+2+3+5=12.
LetN=24*3=48,wherep1=2,p2=3.
TheSmarandachefactorpartitionsof48are(1)N=48(2)N=24*2(3)N=16*3(4)N=12*4(5)N=12*2*2(6)N=8*6(7)N=8*3*2(8)N=6*4*2(9)N=6*2*2*2(10)N=4*4*3(11)N=4*3*2*2(12)N=3*2*2*2*222Definitions:Forsimplicity,wewillusethefollowingabbreviations:1)F(α1,α2,αr)=F'(N),whereN=p1α1p2α2.
.
.
prαr.
.
.
pnαnandpristherthprime.
Inotherwords,p1=2,p2=3,andsoforth.
2)ForthecasewhereNisasquare-freenumber,F(1,1,1,1)=F(1#n).
nonesExamples:F(1#2)=F(1,1)=F'(6)=2,6=2*3=p1*p2.
F(1#3)=F(1,1,1)=F'(2*3*5)=F'(30)=5.
Definition:TheSmarandacheStarFunctionF'*(N)isdefinedasF'*(N)=ΣF'(dr),wheredr|N.
d|NInotherwords,F'*=sumofF'(dr)overallthedivisorsofN.
Example:N=12,thedivisorsare1,2,3,4,6and12.
F'*(12)=F'(1)+F'(2)+F'(3)+F'(4)+F'(6)+F'(12)=1+1+1+2+2+4=11.
Theorem:F'*(N)=F'(Np),(p,N)=1,pisprime.
Proof:BydefinitionF'*(N)=ΣF'(dr),wheredr|N.
d|NConsiderdr,adivisorofN,clearlyNp=dr(Np/dr).
Let(Np/dr)=g(dr),thenN=dr*g(dr)foranydivisordrofN,g(dr)isunique,i.
e.
di=djg(di)=g(dj).
Consideringg(dr)asasingleterm(anentity,notfurthersplitintofactors)intheSFPofN*p,onegetsF'(dr)SFPs.
Eachg(dr)contributesF'(dr)factorpartitions.
23TheconditionpdoesnotdivideNimpliesthatg(di)≠djforanydivisor,becausepdividesg(di)andpdoesnotdividedj.
ThisensuresthatthecontributiontowardsF'(Np)fromeachg(dr)isdistinctandthereisnorepetition.
Summingoverallg(dr)'swegetF'(Np)=∑F'(dr)d|NorF'*(N)=F'(Np)whichcompletestheproofofthetheorem.
TheresultthatF'*(N)=F'(Np),(p,N)=1,pisprimecanbeusedtoprovethatαF(α,1)=ΣP(k).
k=0Toseethis,startwithN=pαp1thenF(α,1)=F'(pα*p1)andfromtheprevioustheoremF'(pα*p1)=F'*(pα)=ΣF'(dr).
d|pαThedivisorsofpαarep0,p1pα,soF'(pαp1)=F'(p0)+F'(p1F'(pα)=P(0)+P(1)+P(2)P(α-1)+P(α)orαF(α,1)=ΣP(k).
k=0nTheorem:F(1#(n+1))=ΣnCrF(1#r)r=0wherenCristhenumberofwaysrobjectscanbeselectedfromasetofnobjectswithoutregardtoorder.
Proof:Bytheprevioustheorem,F'(Np)=F'*(N),wherepdoesnotdivideN.
ConsiderthecaseN=p1p2.
.
.
pn.
WehaveF'(N)=F(1#n)andF'(Np)=F(1#(n+1))aspdoesnotdivideN.
Combiningtheseexpressions,wehaveF(1#(n+1))=F'*(N).
(*)24ThenumberofdivisorsofNoftheformp1p2.
.
.
pr,(containingexactlyrprimes)isnCr.
EachofthenCrdivisorsofthetypep1p2.
.
.
prhasthesamenumberofSFPs,namelyF(1#r).
HencenF'*(N)=ΣnCrF(1#r)(**)r=0From(*)and(**),wehavenF(1#(n+1))=ΣnCrF(1#r)andtheproofiscomplete.
r=0NotethatF(1#n)isthenthBellnumber.
Examples:F(1#0)=F'(1)=1F(1#1)=F'(p1)=1F(1#2)=F'(p1p2)=2F(1#3)=F'(p1p2p3)=5(i)p1p2p3(ii)(p1p2)*p3(iii)(p1p3)*p2(iv)(p2p3)*p1(v)p1*p2*p3ApplyingtheprevioustheoremtoF(1#4)3F(1#4)=ΣnCrF(1#r)r=0F(1#4)=3C0F(1#0)+3C1F(1#1)+3C2F(1#2)+3C3F(1#3)=1*1+3*1+1*5=15F(1#4)=F'(2*3*5*7)=F'(210)=15.
(i)210(ii)105*2(iii)70*3(iv)42*5(v)35*6(vi)35*3*2(vii)30*725(viii)21*10(ix)21*5*2(x)15*14(xi)15*7*2(xii)14*5*3(xiii)10*7*3(ixv)7*6*5(xv)7*5*3*2.
Alongsimilarlines,onecanobtainF(1#5)=52,F(1#6)=203,F(1#7)=877,F(1#8)=4140,F(1#9)=21,147.
Definition:F'**(N)=ΣF'*(dr)dr|NwheredrrangesoverallthedivisorsofN.
IfNisasquare-freenumberwithnprimefactors,wewillusethenotationF'**(N)=F**(1#N).
Examples:F'**(p1p2p3)=F**(1#3)=ΣF'(dr)dr|N=3C0F'*(1)+3C1F'*(p1)+3C2F'*(p1p2)+3C3F'*(p1p2p3)F**(1#3)=1+[3F'(1)+F'(p1)]+3[F'(1)+2F'(p1)+F'(p1p2)]+[F'(1)+3F'(p1)3F'(p1p2)+F'(p1p2p3)]F**(1#3)=1+6+15+15=37.
Aninterestingobservationis(1)F**(1#0)+F(1#1)=F(1#2)orF**(1#0)+F*(1#0)=F(1#2)(2)F**(1#1)+F(1#2)=F(1#3)orF**(1#1)+F*(1#1)=F(1#3)(3)F**(1#5)+F(1#6)=F(1#7)26orF**(1#5)+F*(1#5)=F(1#7).
whichsuggeststhepossibilitythatF**(1#n)+F*(1#n)=F(1#(n+2)).
AstrongerpropositionF'(Np1p2)=F'*(N)+F'**(N)isestablishedinthenexttheorem.
Definition:F'n*(N)=ΣF'(n-1)*(dr)n>1dr|NwhereF'*(N)=ΣF'(dr)dr|NanddrrangesoverallthedivisorsofN.
Theorem:F'(Np1p2)=F'(N)+F'**(N).
Proof:Byprevioustheorem,weknowthatF'(Np1p2)=F'*(Np1).
Letd1,d2,d3,dnbeallthedivisorsofN.
ThedivisorsofNp1wouldbed1,d2,d3,dnd1p1,d2p1,d3p1,dnp1F'*(Np1)=[F'(d1)+F'(d2)F'(dn)]+[F'(d1p1)+F'(d2p1)F'(dnp1)]=F'*(N)+[F'*(d1)+F'*(d2)F'*(dn)].
F'*(Np1)=F'*(N)+F'**(N)(bydefinition)=F'*(N)+F'2*(N).
Whichcompletestheproof.
27Theorem:F'(Np1p2p3)=F'*(N)+3F'2*(N)+F'3*(N).
Proof:Byaprevioustheorem,wehaveF'(Np1p2p3)=F'*(Np1p2).
Letd1,d2,d3,dnbeallthedivisorsofN.
ThedivisorsofNp1p2wouldbed1,d2,d3,dnd1p1,d2p1,d3p1,dnp1d1p2,d2p2,d3p2,dnp2d1p1p2,d2p1p2,d3p1p2,dnp1p2.
Therefore,F'*(Np1p2)=[F'(d1)+F'(d2)+F'(d3)F'(dn)]+[F'(d1p1)+F'(d2p1)F'(dnp1)]+[F'(d1p2)+F'(d2p2)F'(dnp2)]+[F'(d1p1p2)+F'(d2p1p2)F'(dnp1p2)]=F'*(N)+2[F'*(d1)+F'*(d2)F'*(dn)]+SwhereS=[F'(d1p1p2)+F'(d2p1p2)F'(dnp1p2)].
Applyingtheprevioustheorem,wegetF'(d1p1p2)=F'*(d1)+F'**(d1)F'(d2p1p2)=F'*(d2)+F'**(d2).
.
.
F'(dnp1p2)=F'*(dn)+F'**(dn).
Summinguptheseexpressions,wehaveS=F'2*(N)+F'3*(N).
SubstitutingthisvalueofSandalsotakingF'*(d1)+F'*(d2)F'*(dn)=F'2*(N)weget,F'(Np1p2p3)=F'*(N)+2F'2*(N)+F'2*(N)+F'3*(N)28F'(Np1p2p3)=F'*(N)+3F'2*(N)+F'3*(n).
Thiscompletestheproof.
Thisresult,whichisabeautifulpattern,canbefurthergeneralized.
Section5OpenProblemsandConjecturesOntheFactor/ReciprocalPartitionTheoryInthischapter,wepresentsomeopenproblemsandconjecturesrelatedtotheSmarandacheFactorPartitionfunction.
Problems:1)ToderiveaformulaforSFPsofagivenlengthmforN=paqaforanyvalueofa.
2)ToderiveaformulaforSFPsofN=p12p22p32.
.
.
pr23)ToderiveaformulaforSFPsofagivenlengthmofN=p1ap2ap3a.
.
.
pra.
4)Toderiveareductionformulaforpaqaasalinearcombinationofpa-rqa-rforr=0toa-1.
Derivesimilarreductionformulaefor(2)and(3)aswell.
5)Ingeneral,inhowmanywayscananumberbeexpressedastheproductofitsdivisors6)Everypositiveintegercanbeexpressedasthesumofthereciprocalofafinitenumberofdistinctnaturalnumbers.
(ininfinitelymanyways.
).
DefineafunctionRm(n)astheminimumnumberofnaturalnumbersrequiredforsuchanexpression.
7)DetermineifeverynaturalnumbercanbeexpressedasthesumofthereciprocalsofasetofnaturalnumbersthatareinArithmeticProgression.
29(8).
LetΣ1/r≤n≤∑1/(r+1)where∑1/rstandsforthesumofthereciprocalsofthefirstrnaturalnumbersandletS1=∑1/rS2=S1+1/(r+k1)suchthatS2+1/(r+k1+1)>n≥S2S3=S2+1/(r+k2)suchthatS3+1/(r+k2+1)>n≥S3andsoon.
Continuingthissequence,afterafinitenumberofiterationsm,Sm+1+1/(r+km)=n.
Remark:Thevalidityofproblem(6)isdeduciblefromproblem(8).
9).
(a)Thereareinfinitelymanydisjointsetsofnaturalnumbersthesumofwhosereciprocalsisunity.
(b)Amongthesetsmentionedin(a),therearesetswhichcanbeorganizedinanordersuchthatthelargestelementofanysetissmallerthanthesmallestelementofthenextset.
Definition:TheSmarandacheFactorPartitionSequenceisdefinedinthefollowingway:Tn=factorpartitionofn=F'(n).
Forexample,T1=1,T8=3,T12=4etc.
SFPSisgivenby1,1,1,2,1,2,1,3,2,2,1,4,1,2,2,5,1,4,1,4,2,2,1,7,2,.
.
.
,Definition:Fornanaturalnumber,letSbethesmallestnumbersuchthatF'(S)=n.
ThesenumberswillbecalledVedamNumbersandthesequenceformedbytheVedamnumbersistheSmarandacheVedamSequence.
TheSmarandacheVedamSequenceisgivenasfollows:Tn=F'(S)1,4,8,12,16,--,24,.
.
.
30Note:Thereisnonumberwhosefactorpartitionis6,hencethequestionmarkinthatposition.
WewillcallsuchnumbersDullnumbers.
Thereaderisencouragedtoexplorethedistribution,frequencyandotherpropertiesofDullnumbers.
Definition:AnumberissaidtobeaBalunumberifitsatisfiestherelationd(n)=F'(n)=randisthesmallestsuchnumber.
Examples:1,16,36areallBalunumbers.
d(1)=F'(1)=1,d(16)=F'(16)=5,d(36)=F'(36)=9.
EachBalunumber≥16,generatesaBaluClassCB(n)ofnumbershavingthesamecanonicalformsatisfyingtheequationd(m)=F'(m).
Forexample:CB(16)={x|x=p4,pisaprime.
}={16,81,256,SimilarlyCB(36)={x|x=p2q2,pandqareprimes.
}Conjecture:10):ThereareonlyafinitenumberofBaluClasses.
Ifthisistrue,findthelargestBalunumber.
Section6AGeneralResultontheSmarandacheStarFunctionTheorem:nF'(N@1#n)=F'(Np1p2.
.
.
pn)=∑[a(n,m)F'm*(N)]m=0wherema(m,n)=(1/m!
)∑(-1)m-kmCkknk=1Proof:LetthedivisorsofNbed1,d2,dk.
Takethedivisorsof(Np1p2…pn)andarrangethemasfollows:d1,d2,dkcallthesetype0d1pi,d2pi,dkpicallthesetype131d1pipj,d2pipj,d3pipj,dkpipjcallthesetype2.
.
.
.
d1pipj,d2pipj,d3pipj,dkpipjwhereeachtermhastprimes.
Callthesetypet.
.
.
.
.
d1p1.
.
.
pn,d2p1.
.
.
pn,dkp1.
.
.
pn,callthesetypen.
TherearenC0divisorsoftype0.
TherearenC1divisorsoftype1.
.
.
.
TherearenCtdivisorsoftypet.
…TherearenCndivisorsoftypen.
LetNp1p2.
.
.
pn=M.
ThenF*(M)=nC0[Sumofthefactorpartitionsofallthedivisorsofrow0]+nC1[Sumofthefactorpartitionsofallthedivisorsofrow1]+nC2[Sumofthefactorpartitionsofallthedivisorsofrow2]+.
.
.
+nCt[Sumofthefactorpartitionsofallthedivisorsofrowt]+.
.
.
+nCn[Sumofthefactorpartitionsofallthedivisorsofrown].
Considerthecontributionsofthedivisorsetsonebyone.
Row0contributesF'(d1)+F'(d2)F'(dn)=F'*(N).
Row1contributes[F'(d1p1)+F'(d2p1)+F'(d3p1)F'(dkp1)]=[F'*(d1)+F'*(d2)+F'*(d3)F'*(dk)]=F'2*(N).
Row2contributes[F'(d1p1p2)+F'(d2p1p2)F'(dkp1p2)].
ApplyingthetheoremF'(Np1p2p3)=F'*(N)+3F'2*(N)+F'3*(N)oneachoftheterms,F'(d1p1p2)=F'*(d1)+F'**(d1)F'(d2p1p2)=F'*(d2)+F'**(d2).
.
.
32F'(dkp1p2)=F'*(dk)+F'**(dk).
Aftersummingtheserows,wehaveF'2*(N)+F'3*(N).
Atthispoint,wewillwritethecoefficientsintheforma(n,r),forexampleF'(N@1#r)=a(r,1)F'*(N)+a(r,2)F'2*(N)a(r,t)F't*(N)a(r,r)F'r*(N).
Considerrowt,onedivisorsetd1p1p2…pt,d2p1p2…pt,dkp1p2…ptandwehaveF'(d1@1#t)=a(t,1)F'*(d1)+a(t,2)F'2*(d1)a(t,t)F't*(d1)F'(d2@1#t)=a(t,1)F'*(d2)+a(t,2)F'2*(d2)a(t,t)F't*(d2).
.
.
F'(dk@1#t)=a(t,1)F'*(dk)+a(t,2)F'2*(dk)a(t,t)F't*(dk).
Summingupbothsidescolumnwise,wegetforrowtordivisorsoftypetoneofthenCtdivisorsetscontributesa(t,1)F'2*(N)+a(t,2)F'3*(N)a(t,t)F'(t+1)*(N).
Similarlyforrownwehavea(n,1)F'2*(N)+a(n,2)F'3*(N)a(n,n)F'(t+1)*(N).
Allthedivisorsetsoftype0contributenC0a(0,0)F'*(N)factorpartitions.
Allthedivisorsetsoftype1contributenC1a(1,1)F'*(N)factorpartitions.
Allthedivisorsetsoftype2contributenC2[a(2,1)F'2*(N)+a(2,2)F'3*(N)]factorpartitions.
Allthedivisorsetsoftype3contributenC3[a(3,1)F'2*(N)+a(3,2)F'3*(N)a(3,3)F'4*(N)]33AllofthedivisorsetsoftypetcontributenCt[a(t,1)F'2*(N)+a(t,2)F'3*(N)a(t,t)F'(t+1)*(N)]AllofthedivisorsetsoftypencontributenCn[a(n,1)F'2*(N)+a(n,2)F'3*(N)a(n,n)F'(n+1)*(N)]SummingupthecontributionsfromthedivisorsetsofalltypesandconsideringthecoefficientofF'm*(N)form=1to(n+1),wegetthecoefficientofF'*(N)=a(0,0)=a(n+1,1).
ThecoefficientofF'2*(N)=nC1a(1,1)+nC2a(2,1)+nC3a(3,1)nCta(t,1)nCna(n,1)=a(n+1,2).
ThecoefficientofF'3*(N)is=nC2a(2,2)+nC3a(3,2)+nC4a(4,2)nCta(t,2)nCna(n,2)=a(n+1,3).
ThecoefficientofF'm*(N)isa(n+1,m)=nCm-1a(m-1,m-1)+nCma(m,m-1)+nC4a(4,2)nCna(n,m-1).
ThecoefficientofF'n+1*(N)isa(n+1,m+1)=nCna(n,n)=nCn*n-1Cn-1a(n-1,n-1)=nCn*n-1Cn-1.
.
.
2C2a(1,1)=1.
Considera(n+1,2)==nC1a(1,1)+nC2a(2,1)nCta(t,1)nCna(n,1)=nC1+nC2nCn=2n–1=(2n+1–2)/2.
Considera(n+1,3)=34=nC2a(2,2)+nC3a(3,2)nCta(t,2)nCna(n,2)=nC2(21–1)+nC3(22–1)+nC4(23–1)nCn(2n-1–1)=nC2(21)+nC3(22nCn(2n-1)–[nC2+nC3nCn]n=(1/2)[nC2(22)+nC3(23nCn(2n)-∑nCr-nC1-nC0r=0n=(1/2)[∑nCr2r-nC121-nC020]–[2n–n–1]r=0=(1/2)[3n–2n–1]–2n+n+1=(1/2)[3n–2n+1–1]=(1/3!
)[1*3n+1–3*2n+1+3*1n+1–(1)0n+1].
Evaluatinga(n+1,4)a(n+1,4)=nC3a(3,3)+nC4a(4,3)nCna(n,3)=nC3(32+1–23)/2+nC4(33+1–24)/2nCn(3n-1+1–2n)/2=(1/2)[32*nC3+33*nC43n-1*nCn]+(nC3+nC4nCn)-(nC323+nC424nCn2n]n=(1/2)[(1/3)[∑nCr*3r–32*nC2–3nC1-nC0]+r=0n(∑nCr-nC2-nC1-nC0)-(∑nCr*2r–22*nC2–2nC1-nC0)]r=0=(1/2)[(1/3)(4n–9n(n-1)/2–3n–1)+(2n–n(n-1)/2–n–1)–(3n–4n(n-1)/2–2n–1)]a(n+1,4)=(1/4!
)[(1)4n+1–(4)3n+1+(6)2n+1–(4)1n+1+1(0)n+1].
Fromthepatternthatweobserve,itappearsthatthegeneralformulaisra(n,r)=(1/r!
)∑(-1)r-k*rCkknk=035whichwewillproceedtoestablishbyinduction.
Assumethatthefollowingexpressionistrueforrandalln>rr+1a(n,r)=(1/r!
)∑(-1)r-k*rCkkn+1.
k=1Fromthis,thegoalistoderivera(n+1,r+1)=(1/(r+1)!
)∑(-1)r+1-k*r+1Ckkn+1.
k=0Wehavea(n+1,r+1)=nCra(r,r)+nCr+1a(r+1,r)+nCra(r+2,r)nCna(n,r)rr=nCr[(1/r!
)∑(-1)r-krCk*kr]+nCr+1[(1/r!
)∑(-1)r-krCkkr+1]+.
.
.
k=0k=0r+nCn[(1/r!
)∑(-1)r-krCkkn]k=0r=(1/r!
1)r-k*rCk(nCrkr+nCr+1kr+1nCnkn)]k=0rnr-1=(1/r!
1)r-k*rCk(∑nCq*kq-∑nCq*kq)]k=0q=0q=0rnr-1=(1/r!
1)r-k*rCk(1+k)n]-(1/r!
)∑[(-1)r-krCk(∑nCqkq)].
k=0k=0q=0IfwedenotethefirstandsecondtermasT1andT2,wehavea(n+1,r+1)=T1–T2.
ConsiderrT1=(1/r!
)∑[(-1)r-k*rCk(1+k)n]k=036r=(1/r!
)∑[(-1)r-k(r!
/((k!
)(r-k)!
))(1+k)n]k=0r=(1/(r+1)1)r-k((r+1)!
/((k+1)!
(r-k)!
))(1+k)n+1]k=0r=(1/(r+1)1)r-k*r+1Ck+1(1+k)n+1]k=0r=(1/(r+1)1)(r+1)-(k+1)*r+1Ck+1(1+k)n+1]k=0Letk+1=s,wegetfors=1andk=0ands=r+1atk=rr+1=(1/(r+1)!
)∑(-1)(r+1)-2*r+1Cs(s)n+1].
s=1Replacingsbyk,wegetr+1=(1/(r+1)1)(r+1)-k*r+1Ck(k)n+1].
k=1Ifweincludethek=0case,wegetr+1T1=(1/(r+1)1)(r+1)-k*r+1Ck(k)n+1]k=0T1istherighthandsideofther+1formulathatweneedtoderivefromtheinductivehypothesis.
Tocompletetheformula,weneedtoshowthata(n+1,r+1)=T1.
Fromtheexpressiona(n+1,r+1)=T1–T2wehavetoprovethatT2=0.
37rr-1T2=(1/r!
)∑[(-1)r-krCk(∑nCqkq)].
k=0q=0r=(1/r!
)∑[(-1)r-krCk(nC0k0+nC1k1+nC2k2nCr-1kr-1)].
k=0rr=(1/r!
)∑[(-1)r-krCk]+nC1[(1/r!
)∑((-1)r-k*rCkk)]+k=0k=0rrnC2[(1/r!
)∑((-1)r-krCkk2nCr-1[(1/r!
)∑((-1)r-krCkkr-1)]k=0k=0rr=(1/r!
)∑[(-1)r-k*rCk]+nC1[(1/r!
)∑((-1)r-k*rCkk)]+k=0k=0[nC2*a(2,r)+nC3*a(3,r)nCr-1*a(r-1,r)]=X+Y+Z,whererX=(1/r!
)∑[(-1)r-k*rCk]k=0rY=nC1[(1/r!
)∑((-1)r-k*rCkk)]k=0Z=[nC2*a(2,r)+nC3*a(3,r)nCr-1*a(r-1,r)].
WeshallprovethatX=0,Y=0andZ=0.
rX=(1/r!
)∑[(-1)r-k*rCk]k=0r=(1/r!
)∑[(-1)r-k*rCr-k].
38k=0Withthechangeofvariablesr–k=w,wegetk=0,w=randk=r,w=0.
0=(1/r!
)∑[(-1)w*rCw]w=rr=(1/r!
)∑[(-1)w*rCw]w=0=(1–1)r/r!
=0.
Therefore,X=0.
rY=nC1[(1/r!
)∑((-1)r-k*rCkk)]k=0r=nC1[(1/(r-1)1)r-1-(k-1)*r-1Ck-1)]k=1r-1=nC1[(1/(r-1)1)r-1-(k-1)*r-1Ck-1)]k-1=0=nC1[1/(r-1)!
)(1-1)r-1=0.
Therefore,wehaveproventhatY=0.
ToprovethatZ=0webeginwithZ=[nC2*a(2,r)+nC3*a(3,r)nCr-1*a(r-1,r)]andconsiderthematrix39a(1,1)a(1,2)a(1,3)a(1,4).
.
.
a(1,r)a(2,1)a(2,2)a(2,3)a(2,4).
.
.
a(2,r)a(3,1)a(3,2)a(3,3)a(3,4).
.
.
a(3,r)a(4,1)a(4,2)a(4,3)a(4,4)a(4,5).
.
.
a(4,r).
.
.
.
.
a(r-1,r-1)a(r-1,r).
.
.
a(r,1)a(r,2)a(r,3).
.
.
a(r,r-1)a(r,r)Thediagonalelementsareunderlinedandtheelementsabovethemaindiagonalareinbold.
Wehavera(1,r)=[(1/r!
)∑((-1)r-k*rCk*k)]=Y/nC1=0forr>1.
k=0Alloftheelementsofthefirstrowotherthana(1,1)arezero.
Also,a(n+1,r)=a(n,r-1)+r*a(n,r)whichcaneasilybeestablishedbysimplifyingtherighthandside.
Thisgivesusa(2,r)=a(1,r-1)+r*a(1,r)=0forr>2,i.
e.
a(2,r)canbeexpressedasalinearcombinationoftwoelementsofthefirstrow,(otherthantheoneonthemaindiagonal).
Thisimpliesthata(2,r)=0forr>2.
Similarly,a(3,r)canbeexpressedasalinearcombinationoftwoelementsofthesecondrowofthetypea(2,r)withr>3.
Thisimpliesthata(3,r)=0forr>3.
Repeatingthisprocess,wegeta(r-1,r)=0.
Sinceeachofthecoefficientsa(i,j)intheformulaforZarezero,itfollowsthatZ=0.
40Therefore,withX=Y=Z=0,givingT2=0.
Fromthis,a(n+1,r+1)=T1–T2=T1Sincer+1T1=(1/(r+1)1)(r+1)-k*r+1Ck(k)n+1]k=0itfollowsthatr+1a(n+1,r+1)=(1/(r+1)1)(r+1)-k*r+1Ck(k)n+1]k=0whichisther+1expressionoftheinductivehypothesis.
Therefore,theproofofthetheoremiscomplete.
Remark:Thisproofisverylengthyandinvolvesagreatdealofalgebra.
Readersareencouragedtosearchforamoreconciseapproach.
Section7MoreResultsandApplicationsoftheGeneralizedSmarandacheStarFunctionTheorem:αF'n*(pα)=Σn+k+1Cn-1P(α–k)k=0Thefollowingidentitywillbeusedintheproof.
αΣr+k-1Cr-1=α+rCr.
(*)k=0Proof:Byinduction.
Basisstep41Thecasewheren=1hasalreadybeenproven.
αF'1*(pα)=Σ1+k+1C1-1P(α–k)k=0Inductivestep:Assumethattheformulaistrueforalln≤r.
αF'r*(pα)=Σr+k-1Cr-1P(α–k).
k=0StartingwithαF'r+1*(pα)=ΣF'r*(pt)k=0RHS=F'r*(pα)+F'r*(pα-1)+F'r*(pα-2F'r*(p)+F'r*(1).
Fromtheinductivehypothesis,wehaveαF'r*(pα)=Σr+k+1Cr-1P(α–k).
k=0ExpandingtheRHSfromk=0toαF'r*(pα)=r+α-1Cr-1P(0)+r+α-2Cr-1P(1)r-1Cr-1P(α)F'r*(pα-1)=r+α-2Cr-1P(0)+r+α-3Cr-1P(1)r-1Cr-1P(α-1)F'r*(pα-2)=r+α-3Cr-1P(0)+r+α-4Cr-1P(1)r-1Cr-1P(α-2).
.
.
F'r*(p)=rCr-1P(0)+r-1Cr-1P(1)F'r*(1)=r-1Cr-1P(0).
Summinguptheleftandrightsidesseparately,wefindthatLHS=F'(r+1)*(pα).
42TheRHScontainsα+1termsinwhichP(0)occurs,αtermsinwhichP(1)occursandsoon.
Expressingthisasaseriesofsummations:αα-11RHS=[∑r+k-1Cr-1]*P(0)+∑r+k-1Cr-1P(1)r+k-1Cr-1P(α-1)k=0k=0k=00+∑r+k-1Cr-1P(α)k=0Applyingthe(*)identitytoeachofthesummations,wehaveRHS=r+αCrP(0)+r+α-1CrP(1)+r+α-2CrP(2)rCrP(α)α=Σr+kCrP(α).
k=0Or,αF'(r+1)*(pα)=Σr+kCrP(α).
k=0Thepropositionistrueforn=r+1,aswehaveαααF'*(pα)=∑P(α-k)=∑kC0P(α-k)=∑k+1-1C1-1P(α-k).
k=0k=0k=0Hencebyinductionthepropositionistrueforalln,andtheproofiscomplete.
Theorem:n-r∑nCr+kr+kCrmk=nCr(1+m)(n-r).
k=0Proof:n-rLHS=∑nCr+kr+kCrmkk=043n-r=∑(n!
)/((r+k)!
*(n-r-k)!
)*(r+k)!
/((k)!
*(r)!
)*mkk=0n-r=∑(n!
)/((r)!
*(n-r)!
)*(n-r)!
/((k)!
*(n-r-k)!
)*mkk=0n-r=nCr∑n-rCkmkk=0=nCr(1+m)(n-r).
Whichcompletestheproofofthetheorem.
Theorem:nFm*(1#n)=∑nCrmn-rF(1#r)r=0Proof:Byinduction.
Basisstep.
Fromsectionfour,wehavennF*(1#n)=F(1#(n+1))=∑nCrF(1#r)=∑nCr(1)n-rF(1#r)r=0r=0sotheformulaistrueform=1.
Inductivestep.
Assumethattheformulaistrueform=s,whichisnFs*(1#n)=∑nCrsn-rF(1#r).
r=0Or01Fs*(1#0)=∑nC0s0-rF(1#0)Fs*(1#1)=∑nC1s1-rF(1#1)r=0r=04423Fs*(1#2)=∑nC2s2-rF(1#1)Fs*(1#3)=∑nC1s3-rF(1#3)r=0r=0Fs*(1#0)=0C0F(1#0)----(0)Fs*(1#1)=1C0s1F(1#0)+1C1s0F(1#1)----(1)Fs*(1#2)=2C0s2F(1#0)+2C1s1F(1#1)+2C2s0F(1#2)----(2).
.
.
Fs*(1#r)=rC0srF(1#0)+rC1s1F(1#1)rCrs0F(1#r)----(r).
.
.
Fs*(1#n)=nC0srF(1#0)+nC1s1F(1#1)nCns0F(1#r)----(n)MultiplyingtherthequationbynCrandthensumming,theRHSsidebecomes[nC00C0s0+nC11C0s1+nC22C0s2+.
.
.
+nCkkC0sk+.
.
.
+nCnnC0sn]F(1#0)[nC11C1s0+nC22C1s1+nC33C1s2+.
.
.
+nCkkC1sk+.
.
.
+nCnnC1sn]F(1#1).
.
.
[nCrrCrs0+nCr+1r+1Crs1+.
.
.
+nCr+kr+kCrsk+.
.
.
+nCnnCrsn]F(1#r)+[nCnnCns0]F(1#n)nn-r=∑{∑nCr+kr+kCrsk}F(1#r)r=0k=0n=∑nCr(1+s)n-rF(1#n)bytheprevioustheorem.
r=0nLHS=∑nCrFs*(1#r)r=045LetN=p1p2p3.
.
.
pn.
ThentherearenCrdivisorsofNcontainingexactlyrprimes.
ThenLHS=thesumofthesthSmarandachestarfunctionsofallthedivisorsofN,orF'(s+1)*(N)=F(s+1)*(1#n).
Therefore,wehavenF(s+1)*(1#n)=∑nCr(1+s)n-rF(1#n)r=0whichistheinductionexpressionfors+1.
Therefore,theformulaistrueforallnandtheproofiscomplete.
Note:Fromtheprevioussection,wehavenF'(N@1#n)=F'(Np1p2.
.
.
pn)=∑[a(n,m)F'm*(N)]m=0wherema(m,n)=(1/m!
)∑(-1)m-kmCkkn.
k=1IfN=p1p2.
.
.
pk,wehavenkF(1#(k+n)=∑[a(n,m)∑kCtmk-tF(1#t)].
m=0t=0ThisformulaprovidesawaytoexpressBnintermsofsmallerBellnumbers.
Inaway,itisageneralizationofthetheoremnF(1#(n+1)=ΣnCrF(1#r)r=0thatwasproveninsection4.
Theorem:αnF(α,1#(n+1))=∑∑nCrF(k,1#r)k=0r=0Proof:LHS=F(α,1#(n+1))=F'(pαp1p2p3.
.
.
pn+1)=46F'*(pαp1p2p3.
.
.
pn)+∑F'(allthedivisorscontainingonlyp0)+ΣF'(allthedivisorscontainingonlyp1)+ΣF'(allthedivisorscontainingonlyp2)+.
.
.
ΣF'(allthedivisorscontainingonlypr)+.
.
.
ΣF'(allthedivisorscontainingonlypα).
nnnn=∑nCrF(0,1#r)+∑nCrF(1,1#r)+∑nCrF(2,1#r)+∑nCrF(3,1#r)r=0r=0r=0r=0nnnCrF(k,1#r)nCrF(α,1#r)r=0r=0αn=∑∑nCrF(k,1#r).
k=0r=0ThisisareductionformulaforF(α,1#(n+1)).
AResultofSignificanceFromthefirsttheoremofsection4nF'(pα@1#(n+1))=F(α,1#(n+1))=∑a(n+1,m)F'm*(N)m=0wherema(n+1,m)=(1/m!
)∑(-1)m-k*mCk*kn+1k=1andαF'm*(pα)=∑m+k-1Cm-1P(α-k).
k=0Thisisthefirstresultofsomesubstance,givingaformulaforevaluatingthenumberofSmarandacheFactorPartitionsofnumbersthatcanberepresentedina(oneofthemostsimple)particularcanonicalform.
Thecomplexityisevident.
Thechallengingtaskposedtothereaderistoderivesimilarexpressionsforothercanonicalforms.
47Section8PropertiesoftheSmarandacheStarTriangleDefinition:Thefollowingexpressionwasestablishedinsection6.
ma(n,m)=(1/m!
)∑(-1)m-k*mCk*knk=1wehavea(n,n)=a(n,1)=1anda(n,m)=0form>n.
Nowifthetermsarearrangedinthefollowingwaya(1,1)a(2,1)a(2,2)a(3,1)a(3,2)a(3,3).
.
.
a(n,1)a(n,2).
.
.
a(n,n-1)a(n,n)wegetatriangleofnumbersthatwewillcalltheSmarandacheStarTriangle(SST).
ThefirstfewrowsoftheSSTare111131176111525101SomepropertiesoftheSST.
1)Thefirstandlastelementofeachrowis1.
2)Theelementsofthesecondcolumnare2n-1–1,wherenistherownumber.
3)ThesumofalltheelementsinthenthrowisthenthBellnumber.
48Justificationofproperty3:Fromsection6,wehavetheformulanF'(N@1#n)=F'(Np1p2.
.
.
pn)=∑[a(n,m)F'm*(N)].
m=0IfN=1,wehaveF'm*(1)=F'(m-1)*(1)=F'(m-2)*(1)F'(1)=1.
Therefore,nF'(p1p2.
.
.
pn)=Σa(n,m)r=04)Theelementsofarowcanbeobtainedbythefollowingreductionformulaa(n+1,m+1)=a(n,m)+(m+1)*a(n+1,m+1).
5)IfN=pinthetheoremofsection6,wehaveF'm*(p)=m+1.
Therefore,nF'(pp1p2.
.
.
pn)=∑a(n,m)F'm*(N)m=1norBn+1=∑(m+1)a(n,m).
m=16)Theelementsofthesecondleadingdiagonalaretriangularnumbersintheirnaturalorder.
7)Ifpisaprime,pdividesalltheelementsofthepthrowexceptthefirstandthelast,whichareunity.
Thisisestablishedinthefollowingtheorem.
Theorem:a(p,r)≡0(modp)ifpisaprimeand1a(p,r)≡0(modp)andtheproofiscomplete.
Corollary:F(1#p)≡2(modp)Proof:a(p,1)=a(p,p)=1pp-1F(1#p)=∑a(p,k)=∑a(p,k)+2k=0k=2F(1#p)≡2(modp),sincethesummationisevenlydivisiblebyp.
8)Thecoefficientoftherthterm,b(n,r)intheexpansionofxnintheformxn=b(n,1)x+b(n,2)x(x-1)+b(n,3)x(x-1)(x-2)+.
.
.
+b(n,r)xPr+.
.
.
+b(n,n)xPnisequaltoa(n,r).
Theorem:B3n+2isevenandallotherBkisodd.
Proof:Fromsection4,wehaveF'(Nq1q2)=F'*(N)+F'**(N)whenq1andq2areprime50and(N,q1)=(N,q2)=1.
WithN=p1p2p3.
.
.
pn,onecanwriteF'(p1p2p3.
.
.
pnq1q2)=F'*(p1p2p3.
.
.
pn)+F'**(p1p2p3.
.
.
pn)orF(1#(n+2))=F(1#(n+1))+F**(1#n)However,nF**(1#n)=∑nCr2n-rF(1#r)r=0n-1F**(1#n)=∑(nCr2n-rF(1#r))+F(1#n).
r=0ThefirsttermisanevennumberandwewillcallitE.
ThisgivesusF(1#(n+2))-F(1#(n+1))-F(1#n)=E.
Case1:F(1#n)isevenandF(1#(n+1))isalsoevenF(1#(n+2))iseven.
CaseII:F(1#n)isevenandF(1#(n+1))isoddF(1#(n+2))isodd.
Bytheprevioustheorem,F(1#(n+3))-F(1#(n+2))-F(1#(n+1))=E,=>F(1#(n+3))iseven.
FinallywegetF(1#n)isevenF(1#(n+3))iseven.
WeknowthatF(1#2)=2F(1#2),F(1#5),F(1#8)areeven=>B3n+2isevenotherwiseBkisodd.
Section9SmarandacheFactorPartitionsOfaTypicalCanonicalFormInprevioussections,someofthepropertiesoftheSmarandacheFactorPartitionfunctionweredemonstrated.
Inthissection,wewillderiveaformulaforthecasewhere51N=p1α1p2α2.
Theorem:αrα-2jF'(p1αp22)=F(α,2)=∑P(k)+∑∑P(i)k=0j=0i=0wherer=[α/2],α=2rorα=2r+1.
Proof:Theproofwillbebyexaminingasetofmutuallyexclusiveandexhaustivecases.
Onlythenumbersinthebrackets[]aretobefurtherdecomposed.
CaseI:α(p2)[p1αp22]givesF'*(p1α)=∑P(i).
i=0CaseII:{A1}→(p22)[p1αP(α){A2}→(p22p1)[p1α-1P(α-1).
.
.
{Aα}→(p22p1α)[p1α-αP(α-α)=P(0).
Therefore,caseIIcontributesαΣP(i).
i=0CaseIII:{B1}→(p1p2)(p1p2)[p1α-2P(α-2){B2}→(p1p2)(p12p2)[p1α-3P(α-3).
.
.
{Bα-2}→(p1p2)(p1α-1p2)[p1α-αP(α-α)=P(0).
Therefore,caseIIIcontributesα-252ΣP(i).
i=0CaseIV:{C1}→(p12p2)(p12p2)[p1α-4P(α-4){C2}→(p12p2)(p13p2)[p1α-5P(α-5).
.
.
{Cα-4}→(p12p2)(p1α-2p2)[p1α-αP(α-α)=P(0)Therefore,caseIVcontributesα-4ΣP(i).
i=0Note:Thefactorpartition(p12p2)(p1p2)[p1α-3]wascoveredincaseIII.
Therefore,itdoesnotappearincaseIV.
CaseV:{D1}→(p13p2)(p13p2)[p1α-6P(α-4){D2}→(p13p2)(p14p2)[p1α-7P(α-5).
.
.
{Dα-4}→(p13p2)(p1α-3p2)[p1α-αP(α-α)=P(0)Therefore,caseVcontributesα-6ΣP(i).
.
i=0Usingasimilarlineofreasoning,caseVIcontributesα-8ΣP(i).
i=0andwegetcontributionsuptoα-2r53ΣP(i).
i=0where2rcompletestheproof.
Corollary:rF'(p1αp22)=∑(k+2)[P(α-2k)+P(α-2k-1)].
k=0Proof:Intheprevioustheorem,considerthecasewhereα=2r.
2rrα-2jF'(p12rp22)=F(α,2)=∑P(k)+∑∑P(i).
k=0j=0i=0ThesecondtermontheRHScanbeexpandedinthefollowingwayP(α)+P(α-1)+P(α-2)+P(α-3)P(2)+P(1)+P(0)P(α-2)+P(α-3)P(2)+P(1)+P(0).
P(α-4)P(2)+P(1)+P(0).
.
P(2)+P(1)+P(0)P(0).
Summingthetermscolumnbycolumn,=[P(α)+P(α-1)]+2[P(α-2)+P(α-3)]+3[P(α-4)+P(α-5)]+.
.
.
+(r-1)[P(2)+P(1)]+rP(0).
r=∑(k+1)[P(α-2k)+P(α-2k-1)].
54k=0NotethatP(-1)hasbeendefinedtobezero.
Therefore,rrF'(p1αp22)=∑P(k)+∑(k+1)[P(α-2k)+P(α-2k-1)]k=0k=0orrF'(p1αp22)=∑(k+2)[P(α-2k)+P(α-2k-1)].
k=0Inthecasewhereα=2r+1,thesecondtermintheexpressionofthecorollarycanbeexpandedasP(α)+P(α-1)+P(α-2)+P(α-3)P(2)+P(1)+P(0)P(α-2)+P(α-3)P(2)+P(1)+P(0).
P(α-4)P(2)+P(1)+P(0).
.
P(3)+P(2)+P(1)+P(0)P(1)+P(0)Summingthetermsonacolumn-by-columnbasis,wehave[P(α)+P(α-1)]+2[P(α-2)+P(α-3)]+3[P(α-4)+P(α-5)]+.
.
.
+(r-1)[P(3)+P(2)]+r[P(1)+P(0)]r=∑(k+1)[P(α-2k)+P(α-2k-1)],α=2r+1.
k=0Addingonthefirstterm,wegetrF'(p1αp22)=∑(k+2)[P(α-2k)+P(α-2k-1)]k=0Therefore,forallvaluesofα,wehave55[α/2]F'(p1αp22)=∑(k+2)[P(α-2k)+P(α-2k-1)]k=0andtheproofofthecorollaryiscomplete.
Section10Length/ExtentofSmarandacheFactorPartitionsDefinition:IfwedenoteeachSmaradancheFactorPartition(SFP)ofN,sayF1,F2,.
.
.
,FrarbitrarilyandtheFkbetheSFPrepresentationofNastheproductofitsdivisorsinthefollowingway:Fk----N=(h1)(h2)(h3)(h4).
.
.
(ht),whereeachhi(1composed.
Proof:Letθ∈PNandexpressitincanonicalformθ=p1α1p2α2p3α3.
.
.
prαr.
Thend(θ)=(α1+1)(α2+1)(α3+1).
.
.
(αr+1)anditfollowsthatθ∈ΨFkforsomekwhereFkisgivenbyN=(α1+1)(α2+1)(α3+1).
.
.
(αr+1).
Finally,ifθ∈ΨFs,andθ∈ΨFtthenfromuniquefactorizationtheoremFsandFtareidenticalSFPsofN.
Thiscompletestheproof.
62Section13AnAlgorithmforListingtheSmarandacheFactorPartitionsDefinition:Fx'(y)isthenumberoftheSmarandacheFactorPartitions(SFPs)ofywhichinvolvetermsnotgreaterthanx.
IfF1isafactorpartitionofy:F1→x1*x2*x3*.
.
.
xr,thenF1isincludedinFx'(y)iffxi≤xfor1≤i≤r.
Clearly,F'x(y)≤F'(y),andequalityholdsiffx≥y.
Example:F'8(24)=5.
Outof7onlythelast5areincludedinF'8(24).
(1)24(2)12*2(3)8*3(4)6*4(5)6*2*2(6)4*3*2(7)3*2*2*2.
ALGORITHM:Letd1,d2,d3,.
.
.
drbethedivisorsofNindescendingorder.
TolistthefactorpartitionsofN,executethefollowingsequenceofsteps:(A)(1)Startwithd1=Nasthefirstpartition.
(2)Forindexi=2tor(2)(a)Writeallthefactorpartitionsinvolvingdi.
Note:Whenlistingthefactorpartitions,careshouldbetakenthatthetermsfromlefttorightarewrittenindescendingorder.
Atdk≥N1/2≥dk+1,andabove,thiswillensurethatthereisnorepetition.
Example:N=36,Divisorsare36,18,12,9,6,4,3,2,1.
36→3618→18*212→12*3639→9*49*2*26→6*66→6*3*2dk=N1/24→4*3*33→3*3*2*22→NIL1→NILFormulaforF'(N)F'(N)=∑F'dr(N/dr).
dr/NExample:N=216=2333(1)216→F216(1)=1(2)108*2→F108(2)=1(3)72*3→F72(3)=1(4)54*4→F54(4)=2(5)54*2*2(6)36*6→F36(6)=2(7)36*3*2(8)27*8→F27(8)=3(9)27*4*2(10)27*2*2*2(11)24*9→F24(9)=2(12)24*3*3(13)18*12→F18(12)=4(14)18*6*2(15)18*4*3(16)18*3*2*2(17)12*9*2→F12(18)=3(18)12*6*3(19)12*3*3*2(20)9*8*3→F9(24)=5(21)9*6*4(22)9*6*2*2(23)9*4*3*2(24)9*3*2*2(25)8*3*3*3→F8(27)=1(26)6*6*6→F6(36)=4(27)6*6*3*2(28)6*4*3*364(29)6*3*3*2*2(30)4*3*3*3*2*2→F4(54)=1(31)3*3*3*2*2*2→F3(72)=1→F2(108)=0→F1(216)=0.
F'(216)=∑F'dr(216/dr)=31.
dr/NRemarks:ThisalgorithmwouldbequitehelpfulinthedevelopmentofacomputerprogramforthelistingofSFPs.
Section14AProgramForDeterminingtheNumberofSFPsSection13endedwithacommentabouthowthealgorithmwouldbehelpfulinconstructingacomputerprogramtolisttheSFP's.
Inthissection,suchaprograminthe'C'languagethatliststheSFPswillbepresented.
#include/*ThisisaprogramforfindingthenumberoffactorpartitionsofagivennumberwrittenbyK.
Suresh,Softwareexpert,IKOS,NOIDA,INDIA.
*/FILE*f;unsignedlongnp=0;unsignedlongtry_arr[1000];unsignedlongn=0;unsignedlongnum_div=0;unsignedlongdiv_arr[10000];unsignedlongmax_length=0;unsignedlongwidth_arr[1000];unsignedlongextent=0;voidfind_partitions(unsignedlongpos,unsignedlongdiv_idx,unsignedlongprod){unsignedlongi;for(i=div_idx;iCombinatorics:TheoryandApplications",EastWestPressPrivateLtd.
,1985.
10.
AmarnathMurthy,"GeneralizationOfPartitionFunction,IntroducingSmarandacheFactorPartition",SNJ,.
2000.
11.
AmarnathMurthy,"AGeneralResultOnTheSmarandacheStarFunction",SNJ,.
1999.
12.
AmarnathMurthy,"MoreResultsAndApplicationsOfTheGeneralizedSmarandacheStarFunction",SNJ,.
2000.
7113.
AmarnathMurthy,"Length/ExtentofSmarandacheFactorPartitions",SNJ,.
1999.
14.
V.
Krishnamurthy,"Combinatorics:TheoryandApplications"EastWestPressPrivateLtd.
,1985.
15.
AmarnathMurthy,"SomeNewSmarandacheSequences,FunctionsAndPartitions",SNJ,.
2000.
16.
AmarnathMurthy,"SomemoreIdeasonSFPS",SNJ,2000.
72Chapter2SmarandacheSequencesSection1OnTheLargestBaluNumberAndSomeSFPEquationsDefinition:ForapositiveintegerNletd(N)andF'(N)bethenumberofdistinctdivisorsandtheSmarandacheFactorPartitions(SFP)respectively.
IfNisthesmallestnumbersatisfyingd(N)=F'(N)=rforsomer,thenNiscalledaBalunumber.
In[1]MaohuaLeprovesMurthy'sconjecturethatthereareonlyafinitenumberofBalunumbers.
Inthissection,itwillbeprovedthat36isthelargestBalunumber.
Itiswell-knownthatifN=p1a1*p2a2.
.
.
*pkak,thend(N)=(a1+1)(a2+1)(a3+1ak+1).
Proposition1:F'(N)>P(a1)P(a2)Pa3).
.
.
P(ak),fork>1andwhereP(ai)istheadditivepartitionofai.
Proof:From[2]wehavethatF'(pa)=P(a)fork=1.
Weproceedbyinductiononk.
Basisstep:Fork=2,letN=p1a1p2a2.
ConsiderthoseSFPsofNinwhichnoelementisamultipleofp1p2.
InotherwordstheSFPsinwhichthefactorsofp1a1andp2a2areisolated.
ItisquiteevidentthateachSFPofp1a1,whencombinedwitheachSFPofp2a2givesoneSFPofN(=p1a1p2a2).
Therefore,theycontributeP(a1)P(a2)SFPsofN.
TherearemoreSFPslike(p1*p2)*(N/p1*p2)whicharenotcounted.
HencewehaveF'(N)>P(a1)P(a2).
Thepropositionhasbeenestablishedfork=2.
Inductivestep:Letthepropositionbetruefork=m.
Thenwehave73F'(N)>P(a1)P(a2)Pa3).
.
.
P(am).
LetM=N*pm+1am+1.
Then,withF'(pm+1am+1)=P(pm+1am+1)andapplyingsimilarargumentsitisevidentthatF'(M)>P(a1)P(a2)Pa3).
.
.
P(am)P(am+1).
Therefore,bytheprincipleofmathematicalinduction,theproofiscomplete.
WealsohaveP(1)=1,P(2)=2andP(3)=3.
P(4)=5,P(5)=7andwehaveP(a)>a+1fora>4.
ThisgivesusP(a1)P(a2)Pa3).
.
.
P(ak)>(a1+1)(a2+1)(a3+1ak+1)=d(N)forai>4.
Fromtheproposition,wehaveF'(N)>(a1+1)(a2+1)(a3+1ak+1)=d(N)forai>4.
Thenextstepwillbetoproveastrongerproposition.
Proposition2:F(a,b)>F(a)*F(b)fora>2andb>2orinotherwordsF'(p1ap2b)>F'(p1a)*F'(p2b)Proof:Theproofissimilartothatforthefirstproposition.
Proposition3:IfN=p1a1p2a2p3,thenF'(N)>d(N).
Proof:LetM=p1a1p2a2.
Then,from[2]wehaveF'(N)=F'(Mp3)=F'*(M)=∑F'(dr)d/MwheredrM.
ThisequalsF'(p1a1p2a2)+F'(p1a1p2a2-1)+F'(p1a1-1p2a2)+{d(p1a1p2a2)-3}whereeachtermintheexpressionismorethanone.
74Fromproposition2,thisisgreaterthanF'(p1a1)F'(p2a2)+F'(p1a1).
F'(p2a2-1)+F'(p1a1-1).
F'(p2a2)+{d(p1a1p2a2)-3}whereeachtermismorethanone.
ThisisgreaterthanP(p1a1)P(p2a2)+P(p1a1).
P(p2a2-1)+P(p1a1-1).
P(p2a2)+{d(p1a1p2a2)-3}whereeachtermismorethanone.
WehaveP(a)>(a+1)fora>4,fromwhichitfollowsthatthisisgreaterthan(a1+1)(a2+1)+a1(a2+1)+a2(a1+1)+(a1+1)(a2+1)-3.
>2(a1+1)(a2+1),a1,a2>4>2d(M)=d(Mp3)=d(N).
Therefore,wehaveproventhatF('p1a1p2a2p3.
)>d(p1a1p2a2p3.
)fora1,a2>4andtodeterminethelargestBalunumber,weneedtoexamineonlythecaseswherekcompletelistofBalunumbers.
Wealsonotethat1)Theequationd(N)=F'(N)+1hasfivesolutions:2,4,8,24and60.
2)TheequationF'(N)=2*F'(M)hasthreesolutionsF'(6480)=2*F'(2160)=424,F'(360)=2*F'(180)=52,F'(72)=2*F'(30)=16.
Readersarealsoinvitedtofindsolutionsto:(a).
F'(N)=k*F'(M),fordifferentvaluesofk.
(b)F'(N)/F'M)=N/M,(onesolutionisN=360.
M=180.
),anotherisN=210,M=70F'(N)=15.
andF'(M)=5.
Thefollowingtablecontainssomesolutions,whicharehighlightedinbold.
75CanonicalformNumberSFPF'(N)d(N)p4q4r24345=648042450p4q3r24335=216021240p4q2r24325=7209830p4qr24*3*5=2403820p4q24*3=1442910p3q3r23*33*5=108010932p3q2r23*32*5=3605224p3qr23*3*5=1202116p3q23*3=2478p2q2r22*32*5=1802618p2qr22*3*5=601112p2q22*3=1246Pqr2*3*5=3058p2q222*32=3699pq2*3=624p424=1655p323=834p222=423p212p0111Section2SmarandachePascalDerivedSequences76Definition:StartwithanysequenceSb,whichwewillcallthebasesequence.
ASmarandachePascalderivedsequenceSdisdefinedasfollows:nTn+1=ΣrCk*tk+1,wheretkisthekthtermofthebasesequence.
k=0LetthetermsofthebasesequenceSbbeb1,b2,b3,b4,.
.
.
ThentheSmarandachePascalderivedsequenceSd:d1,d2,d3,d4,.
.
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isdefinedasfollows:d1=b1d2=b1+b2d3=b1+2b2+b3d4=b1+3b2+3b3+b4…ndn+1=ΣnCkbk+1k=0Thesederivedsequencesexhibitinterestingpropertiesforsomebasesequences.
Examples:1)Sb:1,2,3,4,naturalnumbers).
Sd:1,3,8,20,48,.
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whichcanbewrittenintheform2x2-1,3x20,4x21,5x22,6x23,.
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ItisalsoeasytoprovethatTn=4(Tn-1-Tn-2)forn>2andTn=(n+1)*2n-2.
2)Sb:1,3,5,7,oddnumbers).
Sd:1,4,12,32,80,.
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77Thefirstdifferencesare1,3,8,20,48,whichisthesameastheSdsequenceforthenaturalnumbers.
TheSdsequencecanalsobewrittenintheform1x20,2x21,3x22,4x23,5x24,.
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Again,itiseasytoprovethatTn=4(Tn-1-Tn-2)forn>2andTn=n*2n-1.
3)SmarandachePascalDerivedBellSequenceStartwiththeSmarandacheFactorPartitions(SFP)sequenceforthesquare-freenumbers,whichisthesameastheBellnumbersequence.
Sb:1,1,2,5,15,52,203,877,4140,.
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ThederivedsequenceisSd:1,1,2,5,15,52,203,877,4140,.
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whichisthesamesequence.
Becauseofthisproperty,wewillcallitthePascalSelfDerivedSequence.
4)StartwiththeFibonaccisequenceasthebasesequenceSb:1,1,2,3,5,8,13,21,34,55,89,114,233,.
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ThederivedsequenceisSd:1,2,5,13,34,89,233,.
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whichiseveryotherelementoftheFibonaccisequence.
Fromthis,itfollowsthatnF2n=ΣnCk*Fk,whereFkisthekthtermofthebaseFibonaccisequence.
k=0IfwetakethepreviousderivedsequenceasthebasesequencewegetthefollowingderivedsequenceSddSdd:1,3,10,35,125,450,1625,5875,21250,.
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Itisinterestingtonotethatthefirsttwotermsaredivisibleby50,thenexttwotermsby51,thenexttwoby52,thenexttwoby53andsoforth.
ExpressedasaformulaT2n≡T2n-1≡0(mod5n).
78Ifwecarryoutthedivision,wehave1,3,2,7,5,18,13,47,34,123,89,andthesequenceformedbytheoddnumberedtermsis1,2,5,13,34,89,.
.
whichistheoriginalsequenceSdthatwasusedasthebase.
Anotherinterestingcharacteristicofthe(*)sequenceisthateveryevennumberedtermisthesumofthetwoadjacentoddnumberedterms.
(3=1+2,7=2+5,18=5+13etc.
).
Conjecture:FnisthenthFibonaccinumber.
2m+1rF2m+1=(1/5m)Σ{2m+1Cr(ΣrCkFk)}.
r=0k=0whichwouldbeyetanotherbeautifulresultinvolvingtheFibonaccinumbersifitcouldbeproven.
Note:AcompanionformulawheretheFibonaccinumbersarereplacedwiththeLucasnumberscouldalsobeconsidered.
ThenextoperationwithbethePascalisationoftheFibonaccisequencewiththeindexesinarithmeticprogression.
ConsiderthefollowingsequenceformedbytheFibonaccinumberswhoseindexesareinarithmeticprogression.
F1,Fd+1,F2d+1,F3d+1,.
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onPascalisationgivesthefollowingsequence1,d*F2,d2*F4,d3*F6,d4*F8dn*F2n,.
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Ford=5,wehavethesequencesBasesequence:F1,F6,F11,F16,.
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1,13,233,4181,46368,.
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Derivedsequence:1,14,260,4920,93200,.
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79inwhichwenoticethat260=20*(14-1),4920=20*(260-14),93200=4920-260),whichsuggeststhefollowingconjecture.
Conjecture:ThetermsofthePascalderivedsequenceford=5aregivenbyTn=20.
(Tn-1-Tn-2)forn>2.
Ford=8wehaveBasesequence:F1,F9,F17,F25,.
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Sb----1,34,1597,75025,.
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Sd----1,35,1666,79919,.
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=1,35,(35-1)*72,(1666-35)*72,.
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etc.
whichsuggeststhefollowingconjecture.
Conjecture:ThetermsofthePascalderivedsequenceford=8aregivenbyTn=49*(Tn-1-Tn-2),n>2.
Wealsoputforwardthefollowingsimilarconjectures.
Ford=10,Tn=90*(Tn-1-Tn-2),(n>2).
Ford=12,Tn=182*(Tn-1-Tn-2),(n>2).
Note:Thereseemstobeadirectrelationbetweendandthecoefficientof(Tn-1-Tn-2)(orthecommonfactor)ofeachterm.
5)SmarandachePascalDerivedSquareSequenceStartwiththesequenceofperfectsquaresSb:1,4,9,16,25,.
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Sd:1,5,18,56,160,432,.
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Whichcanbeexpressedintheform1,5*1,6*3,7*8,8*20,9*48Tn=(n+3)tn-1),wheretristherthtermofPascalderivednaturalnumbersequence.
80ItisalsopossibletoderiveTn=2n-2*(n+3)(n)/2.
6)SmarandachePascalDerivedCubeSequenceStartwiththesequenceofperfectsquaresSb:1,8,27,64,125,.
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Sd:1,9,44,170,576,1792,.
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Inthiscase,wehaveTn≡0(mod(n+1)).
Similarlywecanderivesequencesforhigherpowers,whichcanbeanalyzedforpatterns.
7)SmarandachePascalDerivedTriangularNumberSequenceStartwiththesequenceoftriangularnumbersSb:1,3,6,10,15,21,.
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Sd:1,4,13,38,104,272,.
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8)SmarandachePascalDerivedFactorialSequenceStartwiththesequenceoffactorialnumbersSb:1,2,6,24,120,720,5040,.
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Sd:1,3,11,49,261,1631,.
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WecanverifythatTn=n*Tn-1+Tn-2+1.
Openproblem:ArethereinfinitelymanyprimesinthepreviousSdsequenceStartwiththenaturalnumbersequenceagainSb:1,2,3,4,5,.
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ThecorrespondingderivedsequenceisSd:2*2-1,3*20,4*21,5*22,6*23,.
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Usingthisasthebasesequence,wecangetanotherderivedsequence,whichwedenotebySddorSd2:1,4,15,54,189,648,.
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Whichcanberewrittenas811,4*30,5*31,6*32,7*33.
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Similarly,wecanusethisasthebasesequencetogetthenewderivedsequenceSd3:1,5*40,6*41,7*42,8*43,.
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ThepatternofthefirstfewderivedsequencessuggeststhepatternSdk:1,(k+2)*(k+1)0,(k+3)*(k+1)1,(k+4)*(k+1)2k+r)*(k+1)r-2whichcanbeprovenbyinduction.
Generalization:Wecantakeanyarithmeticprogressionwiththefirsttermaandthecommondifferencebasthebasesequenceandgetthederivedkthordersequencestogeneralizetheaboveresults.
Section3DepascalizationofSmarandachePascalDerivedSequencesandBackwardExtendedFibonacciSequenceIntheprevioussequence,westartedwithabasesequenceSbb1,b2,b3,.
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andthenusedthebasesequencetocreatethederivedsequenceSd.
d1=b1d2=b1+b2d3=b1+2b2+b3d4=b1+3b2+3b3+b4…ndn+1=ΣnCkbk+1k=0Definition:GivenaderivedsequenceSd,theprocessofextractingthebasesequenceSbwillbecalledDepascalization.
TheinterestingobservationisthatthiswillinvolvePascal'striangle,althoughwithadifference.
Itisclearthatb1=d182b2=-d1+d2b3=d1-2d2+d3b4=-d1+3d2-3d3+d4.
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Whichsuggeststhegeneralformulanbn+1=Σ(-1)n+k*nCk*dk+1k=0whichcanbeestablishedbyinduction.
Intheexamplestobegiven,wewillseethatdepascalizedsequencesexhibitsomeinterestingpatterns.
TobeginwithwedefinetheBackwardExtendedFibonacciSequence(BEFS)inthefollowingway.
WestartwiththeFibonaccisequence1,1,2,3,5,8,13,21,34,55,89,144,233,.
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whereF1=1,F2=1,andFn-2=Fn-Fn-1,n>2.
Ifweallowntotakenegativevalues0,-1,-2,.
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andwesubtracttheterms,F0=F2-F1=0,F-1=F1-F0=1,F-2=F0-F-1=-1,.
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theFibonaccisequencecanbeextendedbackwards.
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F-6F-5,F-4,F-3,F-2,F-1,F0,F1,F2,F3,F4,F5,F6,F7,F8,F9,.
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-8,5,-3,2,-1,10,1,1,2,3,58,13,21,34,.
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ThissequencewillbecalledtheFibonacciExtendedBackwardsSequence(FEBS).
DepascalizationoftheFibonacciSequenceTheFibonaccisequenceis1,1,2,3,5,8,13,21,34,55,89,144,233,.
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ThecorrespondingdepascalisedsequenceSd(-1)comesouttobe83Sd(-1)----1,0,1,-1,2,-3,5,-8,.
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NotethatthissequenceisnothingmorethantheextendedFibonaccisequencerotated1800aboutF1andthenthelefttermsomitted.
IfwedepascalizeonemoretimetogetthesequenceSd(-2)1,-1,2,-5,13,-34,89,-233,…ThissequencecanbeobtainedfromtheFibonaccisequenceby:1)Removingevennumberedterms.
2)Multiplyingthealternatetermsby–1.
Conjecture1:IfthefirstrtermsoftheFibonacciSequenceareremovedandtheremainingsequenceisPascalised,theresultingderivedsequenceisF2r+2,F2r+4,F2r+6,F2r+8,.
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whereFristherthtermoftheFibonacciSequence.
Conjecture2:IntheFEBS,ifwetakeTrasthefirsttermanddepascalizetherightsideofit,thenwegettheresultingsequenceastheleftsideofit(lookingrightwards)withTrasthefirstterm.
Asanexample,letr=7,T7=13.
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T-6T-5,T-4,T-3,T-2,T-1,T0,T1,T2,T3,T4,T5,T6,T7,T8,T9,.
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-8,5,-3,2,-1,10,1,1,2,3,58,13,21,34,55,89,.
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depascalisationThedepascalizedsequenceis13,8,5,3,2,1,1,0,1,-1,2,-3,5,-8.
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whichisobtainedbyrotatingtheFEBSaround13(T7)by180andthenremovingthetermsontheleftsideof13.
Readersareencouragedtosearchformorefascinatingresults.
Section4ProofoftheDepascalizationTheorem84Intheprevioussection,theoperationofextractingabasesequencefromaderivedsequencewasdefined.
Knownasdepascalization,itwasstatedthatthegeneraltermofthebasesequence(bi)wasgivenbytheformulanbn+1=Σ(-1)n+k*nCk*dk+1.
k=0Inthissection,itwillbeproventhatthisisindeedthecase.
Theorem:Giventhetermsofthederivedsequenced1=b1d2=b1+b2d3=b1+2b2+b3d4=b1+3b2+3b3+b4…ndn+1=ΣnCkbk+1k=0Thevalueoftheelementsofthebasesequencecanbecomputedusingtheformulanbn+1=Σ(-1)n+k*nCk*dk+1.
k=0Proof:Byinductiononthesubscriptofb.
Basisstep:Ifn=0,then(-1)0=1,0C0=1,sob1=d1.
Inductivestep:Lettheformulabetrueforallsubscriptslessthanorequaltok+1.
Thenwehavebk+1=kC0(-1)k+2d1+kC1(-1)k+1d2kCk(-1)2.
Wealsohavedk+2=k+1C0b1+k+1C1b2k+1Crbr+1k+1Ck+1bk+2whichgives85bk+2=(-1)k+1C0b1-k+1C1b2k+1Crbr+1dk+2.
Substitutingthevaluesofb1,b2,.
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.
intermsofd1,d2,thecoefficientofd1is(-1)k+1C0+(-k+1C1)(-1C0)+(-k+1C2)(2C0)1)r*k+1Cr)(rC0)(-1)k+1(k+1Ck)(kC0)-k+1C0+k+1C1.
1C0-k+1C2*2C01)r*k+1Cr*rC01)k+1*k+1Ck*kC0.
Similarly,thecoefficientofd2isk+1C1*1C1+k+1C2*2C11)r+1*k+1Cr*rC11)k+1*k+1Ck*kC1.
Repeatingtheprocess,thecoefficientofdm+1isk+1Cm*mCm+k+1Cm+1*m+1Cm1)r+m*k+1Cr+m*r+mCm+.
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.
+(-1)k+m*k+1Ck*kCmk-m=Σ(-1)h+1k+1Cm+h*m+hCm.
h=0Whichisequalto(k+1)-mΣ(-1)h+1k+1Cm+h*m+hCm+(-1)k+m*k+1Ck+1*k+1Cm.
h=0Byatheoreminsection7ofchapter1,n-r∑nCr+kr+kCrmk=nCr(1+m)(n-r).
k=0Applyingthistheorem,wehave=k+1Cm{1+(-1)}k+1-m+(-1)k+m*k+1Cm=(-1)k+m.
k+1Cm.
86whichshowsthatthepropositionistruefor(k+2)aswell.
Thepropositionhasalreadybeenverifiedfortheindexequalto1,hencebyinductiontheproofiscomplete.
Inmatrixnotationifwewrite[b1,b2,.
.
bn]1xn*[pi,j]'nxn=[d1,d2,.
.
dn]1xnwhere[pi,j]'nxn=thetransposeof[pi,j]nxnand[pi,j]nxnisgivenbypi,j=i-1Cj-1ifi≤jelsepi,j=0.
Then,wegettheresultIf[qi,j]nxnisthetransposeoftheinverseof[pi,j]nxn,thenqi,j=(-1)j+i*i-1Cj-1.
Wealsohave[b1,b2,.
.
bn]*[qi,j]'nxn=[d1,d2,.
.
dn]where[qi,j]'nxn=thetransposeof[qi,j]nxn.
Section5SmarandacheFriendlyNumbersandAFewMoreSequencesDefinition:Ifthesumofanysetofconsecutivetermsofasequenceequalstheproductofthefirstandthelastnumberoftheset,thenthispairiscalledaSmarandacheFriendlyPairwithrespecttothesequence.
1)SmarandacheFriendlyNaturalNumberPairs:Considerthenaturalnumbersequence1,2,3,4,5,6,7,.
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thentheSmarandachefriendlypairsare(1,1),(3,6),(15,35),(85,204),.
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as3+4+5+6=18=3*615+16+1733+34+35=525=15*35.
Thereexistinfinitelymanysuchpairs.
Thisisevidentfromthefactthatif(m,n)isafriendlypairthensoisthepair(2n+m,5n+2m-1).
2)SmarandacheFriendlyPrimePairs:Startingwiththesequenceofprimenumbers2,3,5,7,11,13,17,23,29,.
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.
wehave2+3+5=10=2*5.
Therefore,(2,5)isafriendlyprimepair.
873+5+7+11+13=39=3*13,so(3,13)isafriendlyprimepair.
5+7+1123+29+31=155=5*31,so(5,31)isafriendlyprimepair.
Similarly(7,53)isalsoaSmarandachefriendlyprimepair.
Inafriendlyprimepair(p,q)wedefineqasthebigbrotherofp.
Openproblem:ArethereinfinitelymanyfriendlyprimepairsOpenproblem:Aretherebigbrothersforeveryprimep3)SmarandacheUnder-FriendlyPair:IfthesumofanysetofconsecutivetermsofasequenceisadivisoroftheproductofthefirstandthelastnumberofthesetthenthispairiscalledaSmarandacheUnder-FriendlyPairwithrespecttothesequence.
4)SmarandacheOver-friendlyPair:IfthesumofanysetofconsecutivetermsofasequenceisamultipleoftheproductofthefirstandthelastnumberofthesetthenthispairiscalledaSmarndacheOver-FriendlyPairwithrespecttothesequence.
5)SmarandacheSigmaDivisorPrimeSequence:Thesequenceofprimespn,whichsatisfythecongruencen-1Σpr≡0(modpn).
r=1Thefirstfewtermsofthesequenceare2,5,71,…since5divides10,and71divides568=2+3+567.
Openproblem:IstheSmarandacheSigmaPrimeSequenceinfiniteConjecture:Everyprimedividesatleastonecumulativesum.
6)SmarandacheSmallestNumberWith'n'DivisorsSequence1,2,4,6,16,12,64,24,36,48,1024,.
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d(1)=1,d(2)=2,d(4)=3,d(6)=4,d(16)=5,d(12)=6,.
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d(Tn)=n,whereTnisthesmallestnumberhavingndivisors.
ItisclearthatTp=2p-1,ifpisprime.
ThesequenceTn+1is2,3,5,7,17,13,65,25,37,49,1025,.
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Conjecture:TheTn+1sequencecontainsinfinitelymanyprimes.
Conjecture:SevenistheonlyMersenneprimeintheTn+1sequence.
88Conjecture:TheTn+1sequencecontainsinfinitelymanyperfectsquares.
7)SmarandacheIntegerPartknSequences(SIPS)(i)SmarandacheIntegerPartπnSequence[π1],[π2],[π3],.
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where[]meanstheintegerpartoftheexpression.
Thefirstfewtermsofthesequenceare3,9,31,97,.
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(ii)SmarandacheIntegerPartenSequence[e1],[e2],[e3],.
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2,7,20,54,148,403,.
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Conjecture:EverySIPScontainsinfinitelymanyprimes.
8)SmarandacheSummableDivisorPairs(SSDP)Thisisasetoforderedpairs(m,n),whered(m)+d(n)=d(m+n),whered(n)isthenumberofdivisorsofn.
Forexample,wehaved(2)+d(10)=d(12),d(3)+d(5)=d(8),d(4)+d(256)=d(260),d(8)+d(22)=d(30),etc.
hence(2,10),(3,5),(4,256),(8,22)areSSPDs.
Conjecture:ThereareinfinitelymanySSPDs.
Conjecture:Foreveryintegermthereexistsandintegermsuchthat(m,n)isanSSDP.
9)SmarandacheProductofDigitsSequenceThenthtermofthissequenceTnisdefinedastheproductofthedigitsofn.
1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0,2,4,6,8,10,12,.
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10)SmarandacheSigmaProductOfDigitsNaturalSequenceThenthtermofthissequenceisdefinedasthesumoftheproductsofallthenumbersfrom1ton.
1,3,6,10,15,21,28,36,45,45,46,48,51,55,60,66,73,81,90,90,92,96,.
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89Hereweconsiderthetermsofthesequenceforsomevaluesofn.
Forn=9wehaveTn=45.
Forn=99wehaveTn=2070=452+45.
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SimilarlywehaveT999=(T9)3+(T9)2+T9=453+452+45=(454-1)/(45-1)=(454-1)/44Thepatternsuggestedbytheprevioussequencecaneasilybeproved.
Thiscanbefurthergeneralizedforanumbersystemotherthanbase10.
Foranumbersystemwithbasebthe(br-1)thtermintheSmarandachesigmaproductofdigitssequenceis2[{b(b-1)/2}r+1-1]/{b2-b-2}Furtherexploration:Thetaskaheadistofindthenthtermintheabovesequenceforanarbitraryvalueofn.
11)SmarandacheSigmaProductofDigitsOddSequence1,4,9,16,25,26,29,34,41,50,52,58,68,82,100,103,112,127,148,.
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Itcanbeprovedthatforn=10r-1,TnisthesumofthertermsoftheGeometricprogressionwiththefirsttermas25andthecommonratioas45.
12)SmarandacheSigmaProductOfDigitsEvenSequence2,6,12,20,20,22,26,32,40,40,44,52,62,78,78,84,96,114,138,.
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Itcanagainbeproventhatforn=10r-1,TnisthesumofthertermsoftheGeometricprogressionwiththefirsttermas20andthecommonratioas45.
OpenProblem:Arethereinfinitelymanycommonmembersinsequences{11}and{12}Section6SomeNewSmarandacheSequences,FunctionsandPartitions1)SmarandacheLCMSequence(SLS)L(n)istheLeastCommonMultiple(LCM)ofthenaturalnumbersfrom1throughn.
ThefirstfewnumbersareSLS→1,2,6,12,60,60,420,840,2520,2520,.
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SmarandacheLCMOddSequence(SLOS)90LO(n)istheLeastCommonMultiple(LCM)ofthefirstnoddnaturalnumbers.
SLOS→1,3,15,105,415,4565,.
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TheSmarandacheLCMEvenSequence(SLES)canbedefinedinasimilarwaySLES→2,4,12,24,240,240.
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ItiseasytoseethatT2n+1(SLES)=2nTn(SLOS),whichisadirectconsequenceofthedefinition.
Additionalideasforexploration:A)IfeachtermoftheSLSisincrementedby1,wegetthenewsequence.
2,3,7,13,61,421,841.
2521,.
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DoesthissequencecontaininfinitelymanyprimesB)Does∞ΣL(n)/n!
existIfitdoes,determinethevalue.
n=1C)Does∞Σ1/L(n)existIfitdoes,determinethevalue.
n=12)DivisorSequencesDefineAn={x|d(x)=n}.
ThenA1={1}A2={p|pisaprime}A3={x|x=p2,pisaprime}A4={x|x=p3orx=p1p2,p,p1,p2areprimes}.
A4→6,8,10,14,15,21,22,26,27,.
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91Wehave∑1/Tn=1forA1andthelimitdoesnotexistforA2.
∞Σ1/L(n)existsandislessthanп2/6forA3asn=1∞Σ1/n2=п2/6.
n=1TheabovelimitexistsforAp,wherepisaprime.
ItisapointforfurtherexplorationtodetermineifthelimitsexistforA4,A6andsoforth.
DivisorMultipleSequenceSDMS={n|n=k*d(n)}.
SDMS→1,2,8,9,12,.
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3)QuadPrimeSequenceGeneratorSQPSG={r|90r+11,90r+13,90r+17,90r+19areallprimes}SQPSG→0,1,2,.
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Openproblem:Arethereinfinitelymanytermsinthissequence4)PrimeLocationSequencesDefinitionP0=sequenceofprimes.
P1=sequenceofprimethprimesP1→3,5,11,17,.
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P2=sequenceofprimeth,primethprime.
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.
Pr=sequenceofprimeth,primeth,.
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rtimes,primesOpenproblem:IfTnisthenthtermofPr,thenwhatistheminimumvalueofrforwhich92∞∑1/Tnexistsn=15)PartitionSequences(i)PrimepartitionThenumberofpartitionsofnintoprimepartsSpp(n)→0,1,1,1,12,2,3,.
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(ii)ThenumberofpartitionsofnintocompositepartsSpc(n)→1,1,1,2,1,3,.
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(iii)DivisorpartitionsThenumberofpartitionsintonumbersthataredivisorsofn.
SPd(n)→1,,1,1,2,1,…(iv)Co-primePartitions(SPcp(n))Thenumberofwaysncanbepartitionedintoco-primeparts.
(v)Non-Co-primePartitions(SPncp(n))Thenumberofwaysncanbepartitionedintononco-primeparts.
(vi)PrimeSquarePartitionsThenumberofwaysncanbepartitionedintoprimesquareparts.
Theseideascouldbegeneralizedtodefinemanymoresuchsequences.
6)CombinatorialSequencesDefineasequenceinthefollowingwayT1=1,T2=2,Tn=sumofalltheproductsoftheprevioustermstakentwoatatime(n>2).
ThissequencewillbeabbreviatedSCS(2)andthefirstfewtermsareSCS(2)=1,2,2,8,48,.
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Thisdefinitioncanbegeneralizedinthefollowingway:StartoutwiththeexplicitvaluesofTj=jbeingassignedtothefirstrterms,thendefineallsubsequentterms93Tn=sumofalltheproductsoftheprevioustermstakenratatime,n>r.
ThissequencewillbeabbreviatedSCS(r).
Another,similarsequencecanbedefinedbyLetTk=kfork=1ton.
Tr=sumofallproductsof(r-1)termsofthesequencetaken(r-2)atatime(r>n).
ThissequencewillbeabbreviatedSCvS.
Forn=2,thefirstfewelementsofthesequenceareT1=1,T2=2,T3=3,T4=17,.
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Openproblem:HowmanyoftheconsecutivetermsofSCS(r)arepairwisecoprimeOpenproblem:HowmanyofthetermsofSCvSareprimes(ii)PrimeproductsequencesSPPS(n)Tn=sumofalltheproductsofprimeschosenfromthefirstnprimestaking(n-1)primesatatime.
ThefirstfewtermsareSPPS(n)→1,5,31,247,2927,.
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T1=1,T2=2+3,T3=2*3+2*5+3*5=31.
T4=2*3*5+2*3*7+2*5*7+3*5*7=247etc.
Openproblem:Howmanyofthenumbersinthissequenceareprime7)φ-SEQUENCE(SφS)={n|n=k*φ(n)}SφS→1,2,4,6,8,12,.
.
.
8)PrimeDivisibilitySequenceSPDS={n|ndividespn+1,pnisthenthprime}94SPDS→1,2,3,4,10,.
.
.
9)DivisorProductSequenceTn=Πdkwheredkisadivisorofn.
Thefirstfewtermsofthissequenceare1,2,3,8,5,36,7,64,27,100,11,1728,.
.
.
Section7SmarandacheReverseAutoCorrelatedSequencesandSomeFibonacciDerivedSmarandacheSequencesDefinition:Leta1,a2,a3,.
.
,beasequence.
WedefineaSmarandacheReverseAuto-CorrelatedSequence(SRACS)b1,b2,b3,.
.
.
inthefollowingway:b1=a21,b2=2a1a2,b3=a22+2a1a3,Orbyapplyingtheformulanbn=Σak*an-k+1.
k=1SuchatransformationwillbecalledtheSmarandacheReverseAutoCorrelationTransformation(SRACT).
Example1:Startingwiththebasesequence1,2,3,4,5,.
.
.
or,expressedanotherway1C1,2C1,3C1,4C1,5C1,.
.
.
TheSRACSis1,4,10,20,35,whichcanberewrittenas3C3,4C3,5C3,6C3,7C3,.
.
.
95WewillcallthissequenceSRACS(1)Takingthisasthebasesequence,wecancomputeSRACS(2)1,8,36,120,330,.
,whichcanberewrittenas7C7,8C7,9C7,10C7,11C7,.
.
.
Takingthisasthebasesequence,wehaveSRACS(3)1,16,136,816,3876,.
.
.
or,expressedanotherway15C15,16C15,17C15,18C15,19C15TheexpressionsofthesequencesusingthenCrnotationsuggeststhefollowingconjecture.
Conjecture:ThesequenceobtainedbyperformingtheSmarandacheReverseAutoCorrelationTransformation(SRACT)ntimesstartingwiththesetofnaturalnumbersisgivenbythefollowing:SRACS(n)h-1Ch-1,hCh-1,h+1Ch-1,h+2Ch-1,h+3Ch-1,.
.
.
whereh=2n+1.
Example2:Usingthetriangularnumbersasthebasesequence1,3,6,10,15,.
.
.
orexpressedanotherway2C2,3C2,4C2,5C2,6C2,.
.
.
TheSRACSis1,6,21,56,126,.
.
.
whichcanberewrittenas5C5,6C5,7C5,8C5,9C5,.
.
.
andwecancallitSRACS(1).
96Usingthisasthebasesequence,wecomputeSRACS(2)1,12,78,364,1365,.
.
.
11C11,12C11,13C11,14C11,15C11,.
.
.
TakingthisasthebasesequencewegettheelementsofSRACS(3)1,24,300,2600,17550,.
.
.
23C23,24C23,25C23,26C23,27C23Thispatternsuggeststhefollowingconjecture.
Conjecture:ThesequenceobtainedbyperformingtheSmarandacheReverseAutoCorrelationtransformation(SRACT)ntimesstartingwiththesetofTriangularnumbersisgivenbySRACS(n).
h-1Ch-1,hCh-1,h+1Ch-1,h+2Ch-1,h+3Ch-1,.
.
.
whereh=3*2n.
Conjecture:GiventhebasesequenceasnCn,n+1Cn,n+2Cn,n+3Cn,n+4Cn,.
.
.
ThesetofelementsofthesequenceSRACS(n)isgivenbyh-1Ch-1,hCh-1,h+1Ch-1,h+2Ch-1,h+3Ch-1,.
.
.
whereh=(n+1)*2n.
SomeFibonacciDerivedSmarandacheSequences1.
SmarandacheFibonacciBinarySequence(SFBS)IntheFibonacciRabbitproblemwestartwithanimmaturepair'I',whichmaturesafteroneseasonto'M'.
Inthesecondseason,thismaturepairbreedsanewimmaturepair,inthethirdseasonthefirstmaturepairbreedsanotherimmaturepairandtheimmaturepairbecomesmature.
Inthefourthseason,theimmaturepairreachmaturityandbothmaturepairsbreedanimmaturepair.
Thiscontinues,whereforanyseason,allimmaturepairsbecomematureandallmaturepairsbreedanimmaturepair.
Whenthisisrepeated,wegetthesequenceI,M,MI,MIM,MIMMI,MIMMIMIM,MIMMIMIMMIMMI,IfwereplaceIby0andMby1,wegetthefollowingbinarysequence970,1,10,101,10110,10110101,1011010110110,…Whenthissequenceisconvertedintotheequivalentdecimalform,wehave0,1,2,5,22,181,5814,.
.
.
WewillcallthissequencetheSFBSandwewillderiveareductionformulaforthegeneraltermsofthesequence.
Fromthebinarypattern,weobserveTn=Tn-1Tn-2{thedigitsoftheTn-2placedtotheleftofthedigitsofTn-1.
}.
AlsothenumberofdigitsinTrisnothingbuttherthFibonaccinumberbydefinition.
HencewehaveTn=Tn-1*2F(n-2)+Tn-2.
Openproblem:HowmanyelementsofthissequenceareprimeOpenproblem:HowmanyelementsofthissequenceareFibonaccinumbers2.
SmarandacheFibonacciProductSequenceTheFibonaccisequenceis1,1,2,3,5,8,.
.
.
StartingwithT1=2,andT2=3andthenusingthegeneralformula,Tn=Tn-1*Tn-2wegetthefollowingsequence2,3,6,18,108,1944,209952,…(*)Inthissequence,whichisdeterminedbythefirsttwoterms,youcanfindtheentireFibonaccisequence.
Thisisclearifyouwritethesequenceinthefollowingway:21,31,2131,2132,2233,2335,2538,.
.
.
Fromthis,itcanbeseenthatTn=2Fn-1*3Fn.
ThisideacanbeextendedbychoosingrtermsinsteadofonlytwoandchangingtherecursivetermtoTn=Tn-1Tn-2Tn-3.
.
.
Tn-rforn>r.
Conjecture:Thesequenceobtainedbyincrementingtheelementsof(*)byonecontainsinfinitelymanyprimes.
Conjecture:Thesequenceobtainedbyincrementingtheelementsof(*)byonedoesnotcontainanyFibonaccinumbers.
98Section8SmarandacheStar(Stirling)DerivedSequencesDefinition:Letb1,b2,b3,.
.
.
beasequenceofnumbersSb,thatwillbethebasesequence.
ThentheSmarandacheStarDerivedSequenceSdisdefinedbyusingthestartriangle111131176111525101inthefollowingway.
d1=b1d2=b1+b2d3=b1+3b2+b3d4=b1+7b2+6b3+b4.
.
.
ndn+1=Σa(m,r)*bk+1k=0wherea(m,r)isgivenbyra(m,r)=(1/r!
)Σ(-1)r-k*rCk*knk=1Forexample:(1)IfthebasesequenceSbis1,1,1,.
.
.
thenthederivedsequenceSdis1,2,5,15,52,inotherwordsthesequenceofBellnumbers.
Tn=Bn.
99(2)IfSb-1,2,3,4,.
.
.
thenSd-1,3,10,37,.
.
.
wehaveTn=Bn+1-BnTheSignificanceoftheabovetransformationwillbeclearwhenweconsidertheinversetransformation.
ItisclearthatthestartriangleisnothingbuttheStirlingNumbersoftheSecondkind.
Considertheinversetransformation.
GiventheSmarandacheStarDerivedSequenceSd,wewishtoretrievetheoriginalbasesequenceSb.
Wegetbkfork=1,2,3,4,…inthefollowingway:b1=d1b2=-d1+d2b3=2d1-3d2+d3b4=-6d1+11d2-6d3+d4b5=24d1-50d2+35d3-10d4+d5Thetriangleofcoefficients1-112-31-611-6124-5035-101ismadeupoftheStirlingnumbersofthefirstkind.
Someofthepropertiesofthistriangleare:1)Thefirstcolumnnumbersare(-1)r-1*(r-1)!
,whereristherownumber.
2)Thesumofthenumbersineachrowiszero.
3)Sumoftheabsolutevaluesofthetermsintherthrowisr!
.
AdditionalpropertiesofthetrianglecanbefoundinthebookbyKrishnamurthy.
ThisprovidesuswitharelationshipbetweentheStirlingnumbersofthefirstkindandthoseofthesecondkind,whichcanbebetterexpressedintheformofamatrix.
100Let[b1,k]1xnbetherowmatrixofthebasesequence.
Let[d1,k]1xnbetherowmatrixofthederivedsequence.
Let[Sj,k]nxnbeasquarematrixoforderninwhichsj,kisthekthnumberinthejthrowofthestartriangle(arrayoftheStirlingnumbersofthesecondkind).
Thenwehave[Tj,k]nxnisasquarematrixoforderninwhichtj,kisthekthnumberinthejthrowofthearrayoftheStirlingnumbersofthefirstkind.
Fromthis,wehave[b1,k]1xn*[Sj,k]'nxn=[d1,k]1xn[d1,k]1xn*[Tj,k]'nxn=[b1,k]1xn.
Whichsuggeststhat[Tj,k]'nxnisthetransposeoftheinverseofthetransposeofthematrix[Sj,k]'nxn.
Readersareencouragedtoconstructalternateproofsbyusingacombinatorialapproachorothertechniques.
Section9SmarandacheStrictlyStaircaseSequenceDefinition:Startingwithanumbersysteminthebaseb,wewilldefineasequenceusingthefollowingpostulate.
1)Numbersarelistedinincreasingorder.
2)Inanynumber,thekthdigitislessthanthe(k+1)stdigit.
Forexample,ifb=6,wehavethesequence1,2,3,4,5,12,13,14,15,23,24,25,34,35,45,123,124,125,134,135,145,234,…Forconvenience,wewritethetermsrowwisewiththerthrowcontainingnumberswithrdigits.
(1)1,2,3,4,5.
{5C1=5numbers}(2)12,13,14,15,23,24,25,34,35,45.
{5C2=10numbers}(3)123,124,125,134,135,145,234,235,245,345.
{5C3=10numbers}(4)1234,1235,1245,1345,2345.
{5C4=5numbers}(5)12345.
{5C5=1number}Thefollowingpropertiesarequiteevidentandareeasytoprove.
**Thespaceisconsideredanumberwithzerodigits.
(1)Thereareb-1Cr(5Crinthiscase)numbershavingexactlyrdigits.
(2)Thereare2b-1(25=32,inthiscase)numbersinthefinitesequenceincludingthespacewhichisconsideredasthelonenumberwithzerodigits.
1013.
ThesumoftheproductofthedigitsofthenumbershavingexactlyrdigitsistheabsolutevalueoftherthterminthebthrowofthearrayoftheStirlingnumbersofthefirstkind.
4.
Thesumofallthesumsconsideredin(3)=b!
-1(6!
-1=719inthiscase).
Openproblem:Deriveanexpressionforthesumofalltherdigitnumbersandthereforeforthesumofthewholesequence.
Openproblem:Ifthenthnumberinthesequencehasindexn,deriveaformulatodeterminetheindexofanynumberinthesequence.
Section10TheSumoftheReciprocalsOftheSmarandacheMultiplicativeSequenceDefinition:Thekthterm(k>2)ofamultiplicativesequencewithinitialtermsm1andm2isthesmallestnumberequaltotheproductoftwopreviousterms.
Forexample,withm1=2andm2=3,thesequenceis2,3,6,12,18,24,36,48,54,72,96,108,144,.
.
.
Inthissection,itwillbeprovedthatthelimitofthesumofthereciprocalsofthetermsofthemultiplicativesequenceexistsforallinitialtermsm1andm2.
Theorem:Thesumofthereciprocalsofthemultiplicativesequencewithinitialtermsm1andm2.
ThesumisS=1/{(m1–1)(m2–1)}+1/m1+1/m2.
Discussion:Considerthesequence2,3,6,12,18,24,36,48,54,72,96,108,144,.
.
.
Itcanbewrittenas2,3,2*3,22*3,2*32,23*3,22*32,24*3,2*33,23*32,.
.
.
Or,foreveryn>2,Tn=2r3sforsomerandspair.
Also,foreveryu∈Nandv∈Nthereexistssomek>2forwhich2u*3v=Tk.
Inanutshell,everytermoftheform2x*3yappearsinthesequenceforallvaluesofxandy.
Onsimilarlinesconsideringthe102generalmultiplicativesequencewithm1andm2asthefirsttwotermsofthesequencewehaveallthetermsofthetypem1r*m2soccurringinthesequencewithr+s>1,(r∈N,s∈N).
Proofoftheorem:Let∞S=∑1/Tn.
n=1ConsidertheproductP=(1/m1+1/m12+1/m131/m2+1/m22+1/m23WehavethesumsofthegeometricserieswithcommonratiocompositionoftheDivisorsofANaturalNumberIntoPairwiseCo-PrimeSetsWithnanaturalnumber,letd1,d2,d3,d4,d5,.
.
.
bethedivisorsofn.
Giventhis,wecouldaskthequestion:Inhowmanywayscanwechooseapairofdivisorswhichareco-primetoeachotherSimilarly,inhowmanywayscanonechooseatriplet,quadrupletandsoforthofdivisorsofnwhicharepairwiseco-primeExamples:LetN=48=24*3.
Thetendivisorsof48are1,2,3,4,6,8,12,16,24,and48.
Thesetofco-primepairswillberepresentedbyD2(48)andthesetofco-primetripletsbyD3(48).
D2(48)={(1,2),(1,3),(1,4),(1,6),(1,8),(1,12),(1,16),(1,24),(1,48),(2,3),(4,3),(8,3),(16,3)}.
TheorderofD2(48)=13.
D3(48)={(1,2,3),(1,3,4),(1,3,8),(1,3,16)}.
TheorderofD3(48)=4.
D4(48)D5(48)D9(48)=D10(48).
104Asanotherexample,considern=30=2x3x5(asquarefreenumber).
The8divisorsof30are1,2,3,5,6,10,15,30.
D2(30)={(1,2),(1,3),(1,5),(1,6),(1,10),(1,15),(1,30),(2,3),(2,5),(2,15),(3,5),(3,10),(5,6)}.
TheorderofD2(30)=13.
NotethatthisiseasilygeneralizedtoO[D2(p1p2p3)]=13,wheretheprimesaredistinct.
D3(30)={(1,2,3),(1,2,5),(1,3,5),(2,3,5),(1,3,10),(1,5,6),(1,2,15)}.
TheorderofD3(30)=7.
D4(30)={(1,2,3,5)}.
TheorderofD4(30)=1.
Openproblem:DeterminetheorderofDr(N).
Inthissection,weconsiderthesimplecaseofnbeingasquare-freenumberforr=2,3andsoforth.
(A)r=2.
Areductionformulawillbederived,followedbyadirectformula.
LetN=p1p2p3.
.
.
pnwherepkisaprimefork=1ton.
WedenoteD2(N)=D2(1#n)forconvenience.
WewillderiveareductionformulaforD2(1#(n+1)).
Letqbeaprimesuchthat(q,N)=1.
ThenD2(Nq)=D2(1#(n+1))andbydefinitionD2(1#n)iscontainedinD2(1#(n+1)).
ThisprovidesuswithO[D2(1#n)]elementsofD2(1#(n+1)).
Consideranarbitrarilychosenelement(dk,ds)ofD2(1#n).
ThiselementwhencombinedwithqyieldsexactlytwoelementsofD2(1#(n+1)).
i.
e.
(qdk,ds)and(dk,qds).
HenceeveryelementofthesetD2(1#n)contributestwoadditionalelementswhen105combinedwiththeprimeq.
Theelement(1,q)hasnotbeenconsideredinthepreviouslymentionedcases,thereforethetotalnumberofelementsofD2(1#(n+1))is3timestheorderofD2(1#n)+1.
Andsoitfollowsthat,O[D2(1#(n+1))]=3*O[D2(1#n)]+1.
(*)ApplyingthisformulafortheevaluationofO[D2(1#4)].
WeknowthatO[D2(p1p2p3)]=O[D2(1#3)]=13,henceO[D2(1#4)]=3*13+1=40.
Example:ThiscanbeverifiedbyconsideringN=2*3*5*7=210,wherethedivisorsare1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,210.
D2(210)={(1,2),(1,3),(1,5),(1,6),(1,7),(1,10),(1,14),(1,15),(1,21),(1,30),(1,35),(1,42),(1,70),(1,105),(1,210),(2,3),(2,5),(2,7),(2,15),(2,21),(2,35),(2,105),(3,5),(3,7),(3,10),(3,14),(3,35),(3,70),(5,6),(5,7),(5,14),(5,21),(5,42),(7,6),(7,10),(7,15),(7,30),(6,35),(10,21),(14,15)}.
Therefore,O[D2(210)]=40.
Theformula(*)canbereducedtoadirectformulabyapplyingsimpleinductiontogetO[D2(1#n)]=(3n-1)/2.
(B)r=3.
Forr=3,wederiveareductionformula.
WehaveD3(1#n)iscontainedinD3(1#(n+1))hencethiscontributesO[D3(1#n)]elementstoD3(1#(n+1)).
LetuschooseanarbitraryelementofD3(1#n)say(a,b,c).
Theadditionalprimeqyields(qa,b,c),(a,qb,c),(a,b,qc)orthreeadditionalelements.
Inthiswayweget3*O[D3(1#n)]elements.
LettheproductofthenprimesbeNandlet(d1,d2,d3,.
.
.
dd(N))beallthedivisorsofN.
ConsiderD2(1#n)whichcontainsd(N)-1elementsinwhichonememberisunity=d1.
Inotherwords,(1,d2),(1,d3)1,dd(N)).
Ifqisplacedasthethirdelementwiththeseasthethirdelementwegetd(N)-1elementsofD3(1#(n+1)).
TheremainingelementsofD2(1#n)yieldelementsthatarealreadycoveredinthepreviousparagraph.
ConsideringtheexhaustivecontributionsfromallthreecasesabovewegetO[D3(1#(n+1))]=4*O[D3(1#n)]+d(N)-1O[D3(1#(n+1))]=4*O[D3(1#n)]+2n-1O[D3(210)]=4*O[D3(30)]8-1O[D3(210)]=4*7+8-1=35.
106Toverify,theelementsarelistedhere.
D3(210)={(1,2,3),(1,2,5),(1,3,5),(1,2,7),(1,3,7),(1,5,7),(1,2,15),(1,2,21),(1,2,35),(1,2,105),(1,3,10),(1,3,14),(1,3,35),(1,3,70),(1,5,6),(1,5,14),(1,5,21),(1,5,42),(1,7,6),(1,7,10),(1,7,15),(1,7,30),(2,3,5),(2,3,7),(2,5,7),(2,3,35),(2,5,21),(2,7,15),(3,5,7),(3,5,14),(3,7,10),(5,7,6),(1,6,35),(1,10,21),(1,14,15)}.
Openproblem:ToobtainadirectformulafromthereductionformulaO[D3(1#(n+1))]=4*O[D3(1#n)]+2n–1.
Regardingthegeneralcase,O[Dr(1#n)],wederiveaninequality.
Let(d1,d2,d3,.
.
.
dr)beanelementofO[Dr(1#n)].
IntroducinganewprimeqotherthantheprimefactorsofNweseethatthiselementinconjunctionwithqgivesrelementsofDr(1#(n+1)).
Inotherwords,(qd1,d2,d3,.
.
.
dr),(d1,qd2,d3,.
.
.
drd1,d2,d3,.
.
.
qdr).
Furthermore,Dr(1#n)iscontainedinDr(1#(n+1)).
HencewegetO[Dr(1#(n+1))]>(r+1)*O[Dr(1#n)].
Findingapreciseformulaisadifficulttaskthatisleftasachallengeforthereader.
Thegeneralcaseisanadditionalchallenge.
Section12OntheDivisorsoftheSmarandacheUnarySequenceDefinition:TheSmarandacheUnarySequenceisdefinedasu(n)=11.
.
.
1,orthedigit'1'repeatedpntimes,wherepnisthenthprime.
Itisnotknownifthissequencecontainsaninfinitenumberofprimes.
LetI(m)=11.
.
.
1=(10m–1)/9.
mtimesThenu(n)=I(pn)andthefollowingpropositionwillbeproven.
Proposition:I(p-1)≡0(modp).
Proof:Clearly,9divides10p-1–1.
FromFermat'slittletheoremifp≥7isaprimethenpdivides(10p-1-1)/9as(p,9)=(p,10)=1.
Therefore,pdividesI(p-1).
Thispropositionwillbeusedtoprovethemainresultofthissection.
107Theorem:Ifdisadivisorofu(n)thend≡1(modpn),foralln>3.
Proof:Letdbeadivisorofu(n)andletd=paqbrc.
.
,wherep,q,andrareprimefactorsofd.
Ifpdividesd,thenpdividesu(n).
Also,pdividesI(p-1)fromtheproposition.
Inotherwords,pdivides(10p-1-1)/9andpdivides(10p-1)/9pdivides(10A(p-1)-1)/9andpdivides(10B.
p-1)/9pdivides(10(A(p-1)-B.
p)/9pdivides10B.
p{(10A(p-1)-B.
p-1)/9}pdivides(10A(p-1)-B.
p-1)/9.
ThereexistAandBsuchthatA(p-1)-B*pn=(p-1,pn).
Aspnisaprimetherearetwopossibilities:(i).
(p-1,pn)=1or(ii).
(p-1,pn)=pn.
Inthefirstcase,from(3)wegetpdivides(10-1)/9orpdividesI,whichisabsurdasp>I.
Therefore,(p-1,pn)=pnorpndividesp-1.
p≡1(modpn)pa≡1(modpn).
Alongsimilarlinesqb≡1(modpn)henced=paqbrc.
.
.
≡1(modpn).
Whichcompletestheproof.
Corollary:Foreveryprimep,thereexistsatleastoneprimeqsuchthatq≡1(modp).
Proof:Sinceu(n)≡1(modpn),andeverydivisorofu(n)is≡1(modpn),thecorollaryholds.
108Section13SmarandacheDualSymmetricFunctionsandCorrespondingNumbersoftheTypeofStirlingNumbersoftheFirstKindItisknownthattherisingfactorial(x+1)(x+2)(x+3).
.
.
(x+n),thecoefficientsofdifferentpowersofxaretheabsolutevaluesoftheStirlingnumbersofthefirstkind.
Letx1,x2,x3,.
.
.
xnbetherootsoftheequation(x+1)(x+2)(x+3).
.
.
(x+n)=0.
Thentheelementarysymmetricfunctionsarex1+x2+x3xn=Σx1,(sumofalltherootstakenoneatatime)x1x2+x1x3+.
.
.
xn-1xn=Σx1x2.
(sumofalltheproductsoftherootstakingtwoatatime)Σx1x2x3…xr=(sumofalltheproductsoftherootstakingratatime).
Inthepreviousexpressions,wehavesummedproducts.
Inthefollowingdefinition,thedualoftheseexpressionswillbedefined,wheretherolesofadditionandmultiplicationareinterchanged.
Definition:TheSmarandacheDualSymmetricfunctionsareformedbytakingtheproductofthesumsinsteadofthesumoftheproducts.
Asanexample,thefollowingisachartforthefourvariablesx1,x2,x3,x4.
Elementrysymmetricfuncions(sumoftheproducts)SmarandacheDualSymmetricfunctions(Productofthesums)x1+x2+x3+x4x1x2x3x4x1x2+x1x3+x1x4+x2x3+x2x4+x3x4(x1+x2)(x1+x3)(x1+x4)(x2+x3)(x2+x4)(x3+x4)x1x2x3+x1x2x4+x1x3x4+x2x3x4(x1+x2+x3)(x1+x2+x4)(x1+x3+x4)(x2+x3+x4)x1x2x3x4x1+x2+x3+x4Forconvenience,thevacuouscaseoftakingtheproductofzerosumsatatimeisdefinedtobeone.
Nowifwetakexr=rintheabovewegettheabsolutevaluesoftheStirlingnumbersofthefirstkind.
Forthefirstcolumn:24,50,35,10,1.
Thecorrespondingnumbersforthesecondcolumnare10,3026,12600,24,1.
ThetriangleoftheabsolutevaluesoftheStirlingnumbersofthefirstkindis10911123161161245035101ThecorrespondingSmarandachedualsymmetrictriangleis1113216606110302612600241Thenextrow(5th)numbersare15,240240,2874009600,4233600,120,1.
Thefollowingpropertiesoftheabovetriangleareevident.
(1)Theleadingdiagonalcontainsunity.
(2)Therthrowelementofthesecondleadingdiagonalcontainsr!
.
(3)Thefirstcolumnentriesarethecorrespondingtriangularnumbers.
Readersareencouragedtofindadditionalrelationshipsbetweenthetwotriangles.
Application:TheSmarandacheDualSymmetricfunctionsgiveusanotherwayofgeneralizingtheArithmetic-GeometricMeanInequality.
Onecaneasilyprovethat:(x1x2x3x4)1/4≤[{(x1+x2)(x1+x3)(x1+x4)(x2+x3)(x2+x4)(x3+x4)}1/6]/2≤[{(x1+x2+x3)(x1+x2+x4)(x1+x3+x4)(x2+x3+x4)}1/4]/3≤{x1+x2+x3+x4}/4.
Thegeneralizationoftheaboveinequalitycanalsobeeasilyestablished.
Section14110OntheInfinitudeoftheSmarandacheAdditiveSquareSequenceDefinition:TheSmarandacheadditivesquaresequenceconsistsofallnumbersthatareperfectsquaresandthesumofthedigitsisalsoaperfectsquare.
Thefirstfewtermsare1,4,9,36,81,100,121,144,169,196,225,324,529,.
.
.
Thequestionwhetherthissequenceisinfinitehasbeenopenandournexttaskistoprovethatitisinfactinfinite.
Theorem:Thesumofthedigitsofthesquareofthenumber'onefollowedbynthree's'isaperfectsquareifn=60t2+76t+24,forsometandisequalto(30t+19)2.
Proof:ConsidertheSmarandachePatternedPerfectSquaresequence169,17689,1776889,177768889,.
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whoserootsequenceis13,133,1333,13333.
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Itisclearthatthissequencefollowsapattern.
PropositionI:Thesquareofonefollowedbynthree'sisequaltoonefollowedby(n-1)seven's,followedbysix,followedby(n-1)eight'sfollowedbyanine.
ProofofpropositionI:ThegeneraltermoftheprevioussquaresequenceisTn=10n+3*(11…ntimes)Tn=10n+3*(10n–1)/9=(1/3)(4*10n-1)(Tn)2=(1/9)(16*102n-8*10n+1).
ThegeneraltermoftherootsequenceisTn=1777.
.
.
6888…9=102n+7*10n+1*(111,(n-1)times)+6*10n+80*(111,(n-1)times)+9=102n+7*10n+1*(10n-1-1)/9+6*10n+80*(10n-1–1)/9+9=(1+7/9)*102n+(-70/9+6+8/9)*10n+(-80/9+9)=(1/9)(16*102n-8*10n+1)whichisthesameas(Tn)2.
ThiscompletestheproofofpropositionI.
Thesumofthedigitsofthistypeofsquarenumberis1+6+9+(n-1)(7+8)=15n+1.
TheonlythingthatremainsistoprovethattheDiophantineequation15n+1=k2hasinfinitelymanysolutions.
111PropositionII:Ifn=60t2+76t+24then15n+1isaperfectsquareforallvaluesoft.
ProofofpropositionII:ConsidertheDiophantineequation15n+1=k2orequivalently15n=(k-1)(k+1).
Letk–1=3rk+1=5sandn=r*s.
Subtractingthefirstfromthesecond,wehave5s=3r+2.
Letr=10t+6,thens=6t+4andn=r*s=(10t+6)(6t+4)=60t2+76t+24.
Thisgivesthevalueofk2=(30t+19)2.
ThecombinationofpropositionIandpropositionIIprovesthetheorem.
Byexaminingadditionalformulas,itmaybepossibletofindadditionalinfinitefamiliesofnumbersintheSmarandachePatternedPerfectSquaresequence.
Section15OntheInfinitudeoftheSmarandacheMultiplicativeSquareSequenceInsection14,theinfinitudeoftheSmarandacheAdditiveSquaresequenceasproven.
Inthissection,thesimilarsequencewithmultiplicationreplacingadditionisexaminedanditisproventhatitcontainsaninfinitenumberofterms.
Definition:TheSmarandachemultiplicativesquaresequenceareallperfectsquareswheretheproductofthedigitsisalsoaperfectsquare.
Thefirstfewtermsofthissequenceare1,4,9,144,289,.
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.
Theorem:Thesquareofthenumber8followedbynthree'sisamemberoftheSmarandachemultiplicativesquaresequenceifnisanoddnumber≥3.
Theproductofthedigitsofthesquareisequalto{2(5r+1)*33}2,wherer=(n-1)/2.
Proof:Startwiththesequence693889,69438889,6944388889,694443888889,.
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.
112whichhastherootsequence833,8333,83333,833333,.
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.
wherethepatternisobvious.
PropositionI:Thesquareof8followedbynthree'sis69followedby(n-2)4's,followedby3,followedby(n-2)8'sfollowedby889.
ProofofpropositionI:ThegeneraltermoftherootsequenceisgivenbyTn=8*10n+3*(111…(n)times)Tn=8*10n+1+3*(10n+1–1)/9=(1/3)(25*10n-1).
Uponsquaring,wehave(Tn)2=(1/9){625*102n-50*10n+1}.
ThegeneraltermofthesequenceofsquaresisTn=69444…3888…889=69*102n+4*10n+2(111…n-2times)+3*10n+1+8000*(111…n-2times)+889=69*102n+4*10n+2*(10n-2-1)/9+3*10n+1+8000*(10n-2-1)/9+889=(69+4/9)*102n+(-400/9+30+80/9)*10n+(-8000/9+889)=(1/9){625*102n-50*10n+1}.
Whichisthesameasthesquareoftherootsequence,thereforetheproofofthepropositioniscomplete.
Proofofthetheorem:Returningtothetheorem,theproductofthedigitsofthenumber(1/9){625*102n-50*10n+1}isP=6*9*4n-2*3*8n-2*8*8*9=2(5n-3)*36.
Pisaperfectsquarewhen5n-3isevenorwhennisodd.
Forexample,n=2r+1,inwhichcaseP=2(10r+2)*36={2(5r+1)*33}2.
Thiscompletestheproofofthemaintheorem.
ReadersareencouragedtofindadditionalinfinitefamiliesofnumbersintheSmarandacheMultiplicativeSquaresequence.
Section16AnotherClassificationoftheOceanofSmarandacheSequencesDefinition:Ifasequenceofnaturalnumberscanbeusedtoexpresseverynaturalnumber113asasumofdistinctnumbersinthesequence,thenitissaidtobeaSmarandacheAccomodativeSequence(SAS).
Example:Thesetofpowersof2:1,2,22,23,242n,.
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isanSASsequence.
Thisisthesequenceusedtorepresentnumbersindigitalcomputers.
Ingeneral,n=∑aktk,whereak=0or1andtkisapowerof2.
AsecondexampleofanSASsequenceisthesequenceofFibonaccinumbers.
ReadersareencouragedtoanalyzetheoceanofSmarandachesequencesforadditionalexamples.
ConjectureI:ThefollowingsequencesfromtheSmarandacheinventorycompiledbyHenryIbstedtin[1]areSmarandacheAccomodative.
Sequencesnumbered6,9,10,11,14,15,93,94,95and123.
Thevalidityoftheconjectureisobviousforsequences93,94,95and123.
Definition:Ifallnaturalnumberscanbeexpressedasthesumordifferencetotermsofasequence,thenthesequenceissaidtobeSmarandacheSemi-Accomodative.
Example:Thesetofpowersof3:1,3,32,333n,.
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isSmarandacheSemi-Accomodative.
Ingeneral,theformulaisn=∑aktk,whereak=-1,0or1andtkisapowerof3.
Example:Thesetoftriangularnumbers1,3,6,10,15,m*(m+1)/2,…isSmarandacheSemi-Accomodative,asn=tn+1–tn.
ConjectureII:PrimenumbersareSmarandacheSemi-Accomodativewithak=-1foratmostonlyonevalueofk.
2,3,5,7,11,13,17,.
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1=3-2,4=7-3,6=13-7=17–11,8=3+5=13–5=19–11,…FurtherGeneralization:GivenasequenceT,ifafinitesetofnumbersA={a1,a2.
.
.
ar}existssuchthateverynaturalnumberncouldbeexpressedasn=∑aktk,ak∈A,tkT,thenthesequenceTisdefinedtobeaccommodativew.
r.
t.
A.
(thetermaccommodativeisusedinthesensethatthesequenceaccommodatesallnaturalnumbersasthelinearcombinationofitstermswithafinitesetofcoefficients.
)Asanexample,forSAS,A={1,0}andforSSAS,A={1,0,-1}.
Note:AlargenumberofSmarandachesequencesforwhicht1=2areaccommodativewiththeexceptionof1.
OpenProblem:ItisanopenproblemtofindsequencesforwhichthesetAexists,andintheeventthatitexiststofindthesmallestone.
114Section17PouringaFewMoreDropsintheOceanofSmarandacheSequencesandSeriesInthissection,somefreshideasonSmarandachesequencesandconjecturesarepresented.
1)SmarandacheForwardReverseSumSequence.
T1=1Tn+1=Tn+R(Tn),whereR(Tn)isthedigitsofTnreversed.
Thefirstfewtermsofthissequenceare:1,2,4,8,16,77,154,605,1111,2222,4444,8888,17776,85547,160165,661166,1322332,3654563,7309126,13528163,49710694,.
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77=16+61,605=154+451,etc.
Conjecture:Thereareinfinitelymanypalindromesinthissequence.
Conjecture:16istheonlysquareinthissequence.
2)SmarandacheReverseMultipleSequence.
Thesequenceofnumbersthataremultiplesoftheirreversals,palindromesareconsideredtrivialandarenotincluded.
Thefirstfewtermsofthissequenceare8712,9801,87912,98901,879912,.
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8712=2178*4.
Pointsofnoteaboutthissequence.
1)Thesequenceisinfinite.
2)Therearetwofamiliesofnumbers,onederivedfrom8712andtheotherfrom9801.
Eachfamilyisconstructedbyplacing9'sinthemiddle.
3)Theconcatenationoftwotermsofthissequencederivedfromthesamefamilyisalsoamemberofthatfamily.
1153)SmarandacheFactorialPrimeGenerator.
T1=1nΠTk+1isaprime,whereTnisthesmallestsuchnumber.
k=1Thefirstfewtermsofthissequenceare:1,2,3,5,7,.
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4)SmarandachePrime-PrimeSequence.
Iftheprimesareplacedinsequence2,3,5,7,11,13,17,19,.
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T1=2,Tn+1=primenumberTnThefirstfewtermsare2,3,5,11,31,.
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SinceT1=2,T2=3thesecondprime,T3=5thethirdprime,T4=11thefifthprime.
5)SmarandacheTriangular-TriangularNumberSequence.
Usingthesequenceoftriangularnumbers1,3,6,10,15,21,28,36,45,.
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.
T1=3,Tn+1=triangularnumberTnThefirstfewtermsare3,6,21,231,26796,.
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.
SinceT1=3,T2=6thethirdtriangular,T3=21thesixthtriangular,T4=231thetwenty-firsttriangular.
6)SmarandacheDivisorsofDivisorsSequence.
T1=3,andTn-1=d(Tn),thenumberofdivisorsofTn,whereTnissmallestsuchnumber.
Thefirstfewtermsofthesequenceare3,4,6,12,72,28.
37,22186*3255where37-1=2186and28-1=255}3,4,6,12,72,559872,22186*3255,.
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.
Thesequenceobtainedbyincrementingeachtermintheabovesequenceby1is1164,5,7,13,73,559873,22186*3255+1,.
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Conjecture:Allofthetermsintheprevioussequencebeyondthefirsttermareprime.
Themotivationbehindthisconjecture:Astheneighboringnumberishighlycomposite(thesmallestnumberhavingsuchagivenlargenumberofdivisors),thechancesofitbeingaprimeareveryhigh.
7)SmarandacheDivisorSum-divisorSumSequences(SDSDS).
Considerthefollowingsequencesinwhicheachtermisthesumofthedivisorsofthepreviousterm:a)1,1,1,1,1,1,.
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b)2,3,4,7,8,15,23,24,52,.
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c)5,6,12,28,56,120,240,744,1920,.
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.
d)9,13,14,24,.
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.
e)10,18,39,56,.
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.
f)11,12,28,.
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.
g)16,31,32,63,104,.
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.
h)17,18,.
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.
i)19,20,42,.
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.
IntheabovesequencesTn=σ(Tn-1),withT1asthegeneratorofthesequence.
Anumberwhichappearsinaprevioussequenceisnottobeusedasagenerator.
Openproblem:Howmanyofthenumberslike12,18,24,28,56…aremembersoftwoormoresequencesOpenproblem:AretherenumbersthataremembersofmorethantwosequencesDefinition:WedefinetheSmarandacheDivisorSumGeneratorSequence(SDSGS)asthesequenceformedby(thegenerators)thefirsttermsofeachoftheabovesequences.
1,2,5,9,10,11,16,17,19,.
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.
Openproblem:IsSDSGSfiniteorinfinite8)SmarandacheReducedDivisorSumPeriodicitySequences.
Inthefollowingsequencesthesumoftheproperdivisorsonlyistakentillthesequence117terminatesat'one'orrepeatsitself.
a)1,1,1,.
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.
b)2,1,.
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.
c)3,1,.
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.
d)4,3,1,.
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.
e)5,1,.
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.
f)6,6,6,.
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.
g)7,1,.
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.
h)12,16,15,8,7,1,…For220,whichisthefirstofapairofamicablenumbers:220,284,220,284,.
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.
Wedefinethelifeofanumberasthenumberoftermsinthecorrespondingsequencetilla'one'isarrivedat.
Forexample,thelifeof2is2andthatof12is6.
Thelifeofaperfectnumberlike6or28orthatofaamicablenumberpairlike(220,284)isinfinite.
Thesameistrueforasociablenumberlikethefivenumberchain12496,14288,15472,14536,14264.
Wecancallthemimmortalnumbers.
Openproblem:Ifnisanyarbitrarynumber,isthereanothernumberkwhoselifeisn9)Tn=smallestprimeoftheformn*k+1,k≥1.
Thefirsttermsofthissequenceare2,3,7,5,11,13,29,17,19,11,23,13,53,29,31,17,…Conjecture:Foreverynthereexistsanumberkcomplementofr.
Thenr+s=10n.
Wehave|r2-s2|=|r–s|*|r+s|=|r–s|*10n.
Therefore,thefirstndigitsofs2arethesameasthatofr2=N.
Also,allthenumbersofthetypek=10x+r,areoftherequiredtypeforx>4n+1.
Thiscompletestheproof.
Example:N=12439729=35272,ten'scomplementof3527=6473.
64732=41899729.
Conjecture:LetNbeanndigitnumber.
Foreveryr,thereexistsanumberksuchthat10ndivideskr-N,ifthereisanrthpower≡N(mod10).
Thisisageneralizationoftheproposition.
Section18SmarandachePythagorasAdditiveSquareSequence1)TheSmarandachePythagorasAdditiveSquareSequence.
TheSmarandachePythagorasAdditiveSquaresequenceisthesetofnumbersthatareperfectsquaresandwherethesumofthedigitsisalsoaperfectsquare.
Thefirstfewtermsofthissequenceare1,4,9,100,.
.
.
Thissequenceisinfinite,ascanbeseenfromthefollowingtheorem.
Theorem:Thesquareofthenumber6*(10n-1)/9,isamemberoftheSmarandachePythagorasAdditiveSquareSequenceifnisgivenby41r2-4r,andthesumofthesquareofthedigitsis(4r-2)2.
Proof:Considerthefollowingsequence6,66,666,6666,66666,666666,thesquaresofthetermsinthissequenceare36,4356,443556,44435556,4444355556,Proposition:Thesquareofthenumberformedfromnsixesisequaltothenumber(n-1)'4's,followedbya'3'followedby(n-1)'5'sfollowedby6.
Proofoftheproposition:Thegeneraltermofthe(*)sequenceisgivenbyTn=6*(10n–1)/9,which,whensquared,gives(Tn)2=(4/9)(102n-2*10n+1).
Thegeneraltermofthe(**)sequenceisgivenby1194*10n+1(111.
.
.
n-1,times)+3*10n+50(111.
.
.
n-1,times)+6=4*10n+1(10n-1–1)/9+3*10n+50(10n-1–1)/9+6=(4/9)*102n+(-40/9+3+5/9)*10n+(-50/9+6)=(4/9)(102n-2*10n+1).
Thisisthesameasthatderivedbysquaringthetermsinthe(*)sequence,sotheproofofthepropositioniscomplete.
LetSbethesumofthesquaresofthedigitsofthegeneraltermofthe(**)sequence.
ThenwehaveS=(n-1)*(42+52)+32+62=41*n+4.
ThiswillbeamemberoftheSmarandachePythagorasAdditiveSquareSequenceif41n+4=k2orequivalently,41n=(k–2)(k+2).
Let41r=k+2,n=r(k-2),thenwehavek=41r-2andn=41r2-4r.
Thiscompletestheproofofthetheorem.
Forr=1,wehaven=37andforr=2,wehaven=148.
Note:Itiseasytoseethatthesumofthedigitsofthegeneraltermis9n.
Usingsimilarreasoning,itispossibletofindadditionalinfinitefamiliesofelementsoftheSmarandachePythagorasAdditiveSquareSequence.
Additionalsequences:1)Smarandache(mth)PowerAdditiveSquareSequenceItisdefinedasasequenceinwhichthetermandthesumofthemthpowerofthedigitsarebothperfectsquares.
(Form=1andm=2wegettheSmarandacheadditivesquaresequenceandSmarandachePythagorasAdditiveSquareSequencerespectively.
)2)Smarandache(mth)PowerAdditivemthpowerSequenceWhenm=3,wegettheSmarandacheadditivecubicsequence.
1,8,125,512,1000,1331,…3)Smarandache(mth)PowerAdditiventhpowerSequence:Whenm=2andn=3weget1,27,216,1000,798005999,.
.
.
10973607685048,.
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.
Alargenumberofopenquestionscanbeformulatedbyexpandingontheseresults.
Section19TheNumberofElementstheSmarandacheMultiplicativeSquareSequenceandtheSmarandacheAdditiveSquareSequencehaveinCommonIntheprevioussection,theSmarandacheMultiplicativeSquaresequence1,4,9,144,289,.
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.
andtheSmarandacheAdditiveSquaresequence1201,4,9,36,81,100,121,144,169,196,225,324,529,.
.
.
weredefined.
Anaturalquestiontoaskis,"Howmanyelementsdothesetwosequenceshaveincommon"Inthissection,wewillprovethatthesequenceshaveaninfinitenumberoftermsincommon.
Theorem:Ifn=4m2–3,thesquareofthenumber'9'followedbyn'6'sfollowedby'9'isamemberoftheSmarandacheMultiplicativeSquaresequenceandtheSmarandacheAdditiveSquaresequence.
Thesumofthedigitsofthisnumberis'36m2'andtheproductofthedigitsis1082*20pwherep=4m2.
Proof:Considerthesequence969,9669,96669,966669,Ifwesquarethetermsofthissequence,weget93896,93489561,9344895561,934448955561,Proposition:Thesquareofthenumber'9'followedbyn'6'sfollowedby'9',isequalto93followedby(n-1)'4'sfollowedby'89'followedby(n-1)'5'sfollowedby'61'.
Proof:Considerthenumbersofthe(*)sequence.
Thenumber9followedbyn6'sfollowedbya9.
Tn=9*10n+1+60*111…+9=9*10n+1+60*(10n–1)/9+9=(1/3)(290*10n+7).
N=(1/3)(29*10n+1+7)N2=(1/9)(841*10(2n+2)+406*10n+1+49).
(***)Considerthenumbersofthe(**)sequence,93followedby(n-1)4's,followedby(n-1)5'sfollowedby61.
Tn=93*10(2n+2)+4*10n+3*(111…(n-1)times)+89*10n+1+500*(111…(n-1)times)+61.
=93*10(2n+2)+4*10n+3(10n-1–1)/9+89*10n+1+500*(10n-1–1)/9+61=(93+4/9)*102n+2+(-400/9+89+5/9)*10n+1+(-500/9+61)=(1/9)(841*10(2n+2)+406*10n+1+49).
Thisisthesameastheelementofthe(***)sequence,whichcompletestheproof.
Thenextstepistocompletetheproofofthetheorem.
121Considerthegeneraltermofthe(**)sequenceTn=93444…89555…61.
LettingSrepresentthesumofthedigitsofTnS=9+3+4(n-1)+8+9+5(n-1)+6+1=9(n+3).
LettingPrepresenttheproductofthedigitsofTnP=9*3*4n-1*8*9*5n-1*6*1=1082*20n-1.
Fromthisformula,ifnisodd,thenPisaperfectsquare.
Usingtheformulaforthesumofthedigits,ifS=9(n+3)=k2then9dividesk2,k=3r,9(n+3)=k2=9r2orn+3=r2.
Sincenisodd,riseven,solettingr=2m,wehaven=4m2–3,whichcompletestheproofofthetheorem.
Section20SmarandachePatternedPerfectCubeSequencesConsiderthesequences10011,100011,1000011,1000011,.
.
.
wheretn=1followedby'n'zerosfollowedby11,or10n+3+11.
Thesequenceformedbythecubesofthenumbersinthissequenceis1003303631331,1000330036301331,1000033000363001331,1000003300003630001331,.
.
.
Theorem:ThetwosequencesaboveformaSmarandachepatternedperfectcubesequence.
Proof:Thenthtermof(1)isgivenbyTn=10n+3+11.
HencethenthtermofcorrespondingcubesequenceisTn3=103n+9+33*103n+6+363*10n+3+1331.
Whichhasthepattern,1followedby(n+1)zeros,followedby33,followedby(n)zeros,followedby363,followedby(n-1)zeros,followedby1331.
Thiscompletestheproof.
122Thefollowingsequencesareexamplesofadditionalcubicrootsequences9,99,999,999,.
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.
729,970299,997002999,999700029999,.
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.
97,997,9997,.
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.
912673,991026973,999100269973,999910002699973,…98,998,9998,99998,.
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.
941192,994011992,999400119992,999940001199992,.
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.
TheproofthatthesesequencesarealsoSmarandachePatternedPerfectCubesequencesissimilartotheonegivensotheyareomitted.
Section21TheSmarandacheAdditiveCubicSequenceIsInfiniteDefinition:ThesequenceofnumbersthatareperfectcubesandwherethesumofthedigitsisalsoaperfectcubeiscalledtheSmarandacheadditivecubicsequence.
Thefirstfewnumbersinthissequenceare1,8,125,512,1000,1331,8000,19683,35937,.
.
.
Theorem:ThecubeofthenumberA(n)givenbyA(n)=10n–1,isamemberofTheSmarandacheadditivecubicsequencewhenn=12k3.
Furthermore,thesumofthedigitsof{A(n)}3isequalto216k3=(6k)3.
Proof:ConsidertheSmarandachepatternedperfectcubealongwithitsrootsequence9,99,999,9999,99999,.
.
.
729,970299,997002999,999700029999,.
.
.
ThegeneralformulasfortheelementsofthesesequencesareTn=10n-1,Tn3=103n-3*102n+3*10n-1.
Thecubesequenceisthentn=102n+1*(10n-1–1)+7*102n+2*10n+(10n–1).
Which,onsimplificationgives123tn=103n-3*102n+3*10n-1=Tn3.
Thesumofthedigitsfortn=9(n-1)+7+2+9n=18n.
Ifn=12k3,then,sumofdigitsfortn=216k3=(6k)3.
Thiscompletestheproofofthetheorem.
Section22MoreExamplesandResultsOntheInfinitudeofCertainSmarandacheSequences1)MaohuaLehasgivensomeexamplesthatprovethatthereducedSmarandacheSquare-Digitalsubsequenceisinfinite.
Here,wegiveanotherexample.
ThesquareofthenumbersA(n)=(10n–3)forallvaluesofnyieldstermsofareducedSmarandacheSquare-Digitalsubsequence.
A(n):7,97,997,9997,99997,.
.
.
(A(n))2:49,9409,994009,99940009,9999400009,.
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.
2)TheSmarandachemultiplicativesquaresequencewasdefinedinsection15anditwasproventhatthesequenceisinfinite.
Thefollowingisanotherinfinitefamilyofmembersofthissequence.
A(n):38,338,3338,33338,.
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.
(A(n))2:1444,114244,11142244,1111422244,.
.
ThegeneralformulasfortheelementsofthesesequencesareA(n)=10*(10n-1)/3+8{A(n)}2=10n+2*(10n–1)/9+4*10n+1+(200/9)(10n-1–1)+44={10*(10n-1)/3+8)2.
Theproductofthedigitsof(A(n))2=2n+5,whichisaperfectsquareforoddn(n=2k+1).
Note1:ItisinterestingtoseethatthereverseoftheelementsofA(n)exhibitsthesameproperty.
83,833,8333,83333,.
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.
6889,693889,69438889,6944388889,694443888889.
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.
124Note2:Thesumofthedigitsof(A(n))2isgivenby12+n+2(n-1)=3n+10.
Ifwelookforvaluesofnwhichmake3n+10aperfectsquarer2,wewillgetaninfiniteadditivesquaresequence.
Inotherwords,ifn=(3k2±2k–3)then3n+10=9k2±6k+1=(3k±1)2,isaperfectsquare.
Therefore,forn=(3k2±2k–3),wegetaninfiniteSmarandacheadditivesquaresequence.
Moreover,ifwealsohavek=2m,anevennumber,thenthesenumbersarealsomembersofamultiplicativesquaresequence.
WehavefinallyarrivedataninfinitesequenceofnumbersthataresimultaneouslymembersofboththeSmarandacheadditivesquaresequenceandtheSmarandacheMultiplicativesquaresequence.
Themembersofthesequence{A(n)}2={10*(10n-1)/3+8)}2aresimultaneouslymembersofboththeSmarandacheadditivesquaresequenceaswellasSmarandacheMultiplicativesquaresequenceforn=5,13,37,53,93,117,.
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.
Section23SmarandacheSymmetric(Palindromic)PerfectPowerSequencesDefinition:TheSmarandacheSymmetricPerfectmthPowersequenceisthesetofnumberswhicharesimultaneouslymthpowersandpalindromic.
ThefirstfewtermsoftheSmarandachesymmetricperfectsquaresequenceare:1,4,9,121,484,14641,.
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.
ThefirstfewtermsoftheSmarandacheSymmetricPerfectcubesequenceare:1,8,343,1331,.
.
.
1367631,.
.
.
Inthissection,wewillverifythatthesequenceisinfiniteforsomevaluesofm.
Theorem:TheSmarandachesymmetricperfectmthpowersequencehasinfinitelymanytermsform=1,2,3and4.
Proof:Wewillstartwithm=2.
Considerthefollowingsequences,wherethesecondisthesquaresofthefirst.
11,101,1001,10001,100001,.
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.
121,10201,1002001,100020001,.
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.
TheformulasforthegeneraltermsofthesequencesareTn=10n+1,andTn2=102n+2*10n+1.
Thisverifiesthetheoremform=2.
125Considerthefollowingsequences,wherethesecondisthecubesoftheelementsofthefirst.
11,101,1001,10001,100001,.
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.
1331,1030301,1003003001,1000300030001,.
.
.
Thisverifiesthetheoremform=3.
Considerthefollowingsequences,wherethesecondismadeupofthefourthpowersofthefirst11,101,1001,10001,100001,.
.
.
14641,104060401,1004006004001,.
.
.
Thisverifiesthetheoremform=4.
Note1:Thethreesequencesofsquares,cubesandfourthpowersintheprevioustheoremarealsoexamplesofinfiniteadditivesquare,cubicandfourthpowersequences.
Note2:ThegeneraltermoftherootsequencecanalsobeexpressedintheformTn=2(10n+1).
Note3:Therootsequencecanbetakenas102n+10n+1or103n+102n+10n+1,form=2.
Conjecture:TheSmarandacheSymmetricPerfectmthpowersequencehasinfinitelymanytermsforallvaluesofm.
Section24SomePropositionsOntheSmarandachen2nSequenceDefinition:TheelementsoftheSmarandachen2nsequencearecreatedbyconcatenatingnand2ntogether.
Thefirstfewtermsofthesequenceare12,24,36,48,510,612,714,816,1224,1326,1428,.
.
.
Thenthtermisgivenbytheformulaa(n)=2*n+n*10r,wherer=d(2n),thenumberofdigitsofn.
IthasbeenconjecturedbyRussothatthesequencecontainsinfinitelymanyprimes.
However,ithasbeenproventhatitcontainsnoprimes.
Inthissection,somepropertiesofthesequencewillbepresented.
126Tocomputethefinaldigitsumofanumberthedigitsaresummed.
Ifthatsumisgreaterthanten,thedigitsofthesumareaddedandthisprocessisrepeateduntilasingledigitnumberiscomputed.
ConsiderthefinalsumofthedigitsoftheelementsoftheSmarandachen2nsequence.
Thepatternthatwegetis3,6,9,3,6,9,3,6,9,3,6,9,.
.
.
Bydefinition,thefollowingpropertiesareevident.
(i)Thesumofdigitsisdivisiblebythreeandhenceeachtermofthesequenceisdivisibleby3.
(ii)Theonlyvaliddigitssumoccurringis9,anecessaryconditionforaperfectsquare.
(iii)Thedigitsequence{3,6,9},isrepeatedperiodically.
(iv)Thenthtermisdivisibleby2n.
Thefinalsumofthedigitsforthenthtermisgivenbyd=3,6or9accordinglyasn=3r+1,3r+2,or3r.
LetthesequenceobtainedbydividingthetermsoftheSmarandachen2nsequenceby3becalledtheSmarandachen2nbythreesequence,whichis4,8,12,16,170,204,238,272,.
.
.
Ournextstepwillbetoprovethatthissequencecontainsinfinitelymanyperfectsquares.
Theorem:Thenthtermoftheabovesequenceisaperfectsquareequalton2itself,ifnisgivenbyn={10k+2}/3.
Proof:Wehaven={10k+2}/3,whichhasexactlykdigits.
WegetthecorrespondingtermoftheSmarandachen2nsequencebya(n)=2*n+n*10r=2*{10k+2}/3+[{10k+2}/3]*{10k}.
Therefore,thecorrespondingtermoftheSmarandachen2nbythreesequenceisgivenbyb(n)=a(n)/3=(1/3)*[2*{10k+2}/3+{(10k+2)/3}*(10k)]b(n)=(1/9)*{102k+4*10k+4}={(10k+2)/3}2=n2.
Thiscompletestheproof.
Thesequenceofnumbersis12734,334,3334,3334,.
.
.
andthecorrespondingtermsoftheSmarandachen2nbythreesequenceare1156,111556,11115556,1111155556,.
.
.
(Anothersequenceofpatternedperfectsquares.
)ThecorrespondingtermsoftheSmarandachen2nsequenceare3468,334668,33346668,3333466668,.
.
.
Note:ItfollowsdirectlyfromthistheoremthattheSmarandachen2nsequencecontainsinfinitelymanytermsoftheform3*n2.
Itisthatterma(n)oftheSmarandachen2nsequenceisdivisibleby6n.
WewilldefinetheSmarandachen2nby6nsequencebyc(n)=a(n)}/(6n).
Thefirstfewtermsofthesequenceare2,2,2,2,17,17,17,.
.
.
17,167,167,.
.
.
1667,1667,.
.
Andforsomespecifictermsc(5)=510/30=17,c(49)=4998/294=17,c(50)=50100/300=167,c(499)=499998/(6*499)=167,c(500)=5001000/3000=1667.
Conjecture:Thesequencec(n)containsinfinitelymanyprimes.
Section25TheSmarandacheFermatAdditiveCubicSequenceDefinition:TheSmarandacheFermatAdditiveCubicSequenceisconstructedfromthenumbersthatareperfectcubesandthesumofthecubesoftheirdigitsisalsoaperfectcube.
ThenameofFermatisincludedinthedescriptiontorelateitwiththefactthatthoughthesumoftwocubescannotyieldathirdcube,thesumofmorethantwocubescanbeathirdcube(33+43+53=63).
Thefirstfewtermsofthesequenceare1,8,474552,27818127,.
.
.
where474552=783,and43+73+43+53+53+23=729=9327818127=3033,sumofcubesofdigits=1728=123.
128Theorem:TheSmarandacheFermatAdditiveCubicsequencecontainsaninfinitenumberofterms.
Proof:Considerthefollowingsequence3033,30033,300033,.
.
.
27818127,27081081027,27008100810027,.
.
.
ThegeneraltermisgivenbyTn=27*(10n+1+1)3=27*(103(n+1)+3*102(n+1)+3*10(n+1)+1).
Thesumofthecubesofthedigitsforeverytermis2*(13+23+73+83)=1728=123andtheproofiscomplete.
Note1:Itisinterestingtonotethatthedigits1,2,7,8areusedtwiceineverytermwiththerestofthedigitsbeingzero.
Thesumofthecubesofthedigitsalsoismadefromthesamedigits.
Note2:Apermutationofthefourdigits2178hasthepropertythat4*2178=8712,thenumberobtainedbyreversingthedigits.
Infactthenumbersobtainedbyplacingitadjacenttoitselfanynumberoftimesalsohavethesameproperty.
Inotherwords,4*21782178=87128712.
Thisalsoholdsifanynumberofninesareplacedinthecenter,4*21978=87912,4*219978=879912.
Theonlyothersuchnumberis1089.
(9*1089=9801).
Inthepreviousparagraphs,wehavedescribedasequenceinwhichthesumofcubesofthedigitsisthesameforeachterm.
Wewillnowdescribeasequenceinwhichthesumofthecubesofthedigitsgivesdifferentcubesfordifferentterms.
Theorem:Ifkisapositiveintegerthenthenumber(10n+2-4)3isamemberoftheSmarandacheFermatAdditiveCubicSequencewhenniscanbeexpressedintheform[4*{(103k–1)/27}–1].
Thesumofthecubesofthedigitswillthenequal(6*10k)3.
Proof:Considerthefollowingpatternedperfectcubesequence9963,99963,999963,.
.
.
wherethesequenceofcubesis129988047936,998800479936,999880004799936,.
.
Thegeneraltermoftheoriginalsequenceisgivenby(10n+2–4),whichuponbeingcubed,becomesTn=(10n+2–4)3=103(n+2)-12*102(n+2)+48*10(n+2)–64.
Thenthtermofthesequenceofcubesisgivenby(10n-1)*102n+6+88*102n+4+47*10n+2+(10n-1)*102+36.
WhichcanbereducedtothesequenceTn.
ThesumofthecubesofthedigitsofTniss(d)=2n*93+2*83+43+73+33+63=1458n+1674.
Ifs(d)isacubesayr3,thenwehaver3=1458n+1674orr3=27(54n+62).
Ifr=3sthenwehave27s3=27(54n+62)ors3=(54n+62)=54(n+1)+8,assiseven.
Lets=2u,then8u3=54(n+1)+8or8(u3–1)=54(n+1)orn=4(u3–1)/27-1andnwouldbeanintegerif27dividesu3-1.
Wehave999=27*37=103-1and(103-1)divides(103k–1).
Therefore,27divides(10k)3–1,givingu=10kforallvaluesofk.
Expressedanotherway,n=[4*{(103k–1)/27}–1].
Nowr3=1458n+1674=1458[4(103k–1)/27-1]+1674=216*103k={6*10k}3.
Thiscompletestheproof.
Exploringsimilarsequencesforhigherpowerscanextendthisidea.
Openproblem:Isthesequenceofadditivefourthpowersinfinite130Openproblem:IsthesequencewherethetermsaswellasthefourthpowersofthedigitsarefourthpowersfiniteorinfiniteSection26TheSmarandachenn2SequenceContainsNoPerfectSquaresDefinition:TheSmarandachenn2sequenceisthesetofnumbersformedbyconcatenatingnandn2.
Thefirstfewtermsofthissequenceare11,24,39,416,525,636,749,864,981,10100,11121,12144,.
anditisclearthatthegeneraltermofthesequenceisgivenbya(n)=n*10r+n2,wherer=d(n2),thenumberofdigitsofn2.
Ithasbeenconjecturedinthattherearenoperfectsquaresinthissequence.
Inthissection,weproveatheoremverifyingthatconjecture.
Theorem:ThenecessaryconditiononnthatgivesaperfectsquaretermoftheSmarandachenn2sequenceisa)n≡8or0(mod9).
b)Inthecasewheren=9m,misnotasquarefreenumber.
Proofof(a):Definition:Anumberdissaidtobeavaliddigitssumifd≡1(mod3),ord≡0(mod9).
Proposition:Thedigitssumofaperfectsquareisnecessarilyavaliddigitssum.
Proofofproposition:Considerthesquaresofnumbers1through9,1,4,9,16,25,36,49,64,81,wherethedigitssumsare1,4,9,7,7,9,4,1,9,allofwhichareavaliddigitssum.
Itcanalsoprovedusingthepropertiesofcongruencethatthedigitssumoftheproductoftwonumbersistheproductofthedigitssums.
Therefore,theproofofthepropositioniscomplete.
ThefollowingsequenceisconstructedfromthefinaldigitsumoftheelementsoftheSmarandachenn2sequence.
2,6,3,2,3,3,2,9,9,2,6,3,2,3,3,2,9,9,2,6,3,2,3,3,2,9,9,2.
131Thefollowingtwopropertiesareeasytoverify.
i)Thesequenceofdigits{6,3,2,3,3,2,9,9,2}repeatsperiodically.
ii)Theonlyvaliddigitsumis9anditoccursonlyforn≡8or0(mod9).
Proofof(b):Leta(n)=k2beaperfectsquareinthesequence.
Thenwehavea(n)=k2=n*10r+n2.
Ifn=9m,wheremisasquarefreenumber,thenk2–n2=n*10ror(k+9m)(k-9m)=9m*10r=m*32*2r*5r.
Now,itisevidentthataprimedivisorofmdivideseitherk+9mork-9m,andineithercasemmustdividek.
Ifk=p*m,thentheexpressioncanbewrittenas(p*m+9m)(p*m-9m)=9m*10r=m*32*2r*5r,or(p+9)(p-9)=9*10r=*32*2r*5r.
Onsimilarlines3dividespgivingthatpcanbeexpressedintheformp=3*q,givingus(3q+9)(3q-9)=32*2r*5ror(q+3)(q-3)=3*2r*5r.
Ifwealsohavethatq=3*s,wehave(s+1)(s-1)=2r*5r,ors2=10r+1.
Thedigitsumoftherightmemberis2,whichisnotavaliddigitsumforaperfectsquare.
Thereforetheexpressionhasnosolutioninintegers.
Also,forevenr=2t,thetwoconsecutivenumbers102tand102t+1cannotbothbeperfectsquares.
Thiscompletestheproofofpart(b)ofthetheorem.
Definition:TheReducedSmarandachenn2sequenceisgivenbyb(n)=a(n)/n.
Thefirstfewtermsofthesequenceare11,12,13,104,105,106,107,108,109,1010,1011,1012,1013,.
.
.
Conjecture:ThereareinfinitelymanyprimesinthereducedSmarandachesequence.
Section27PrimesIntheSmarandachennmSequenceDefinition:Form>0,theSmarandachennmsequenceisformedbyconcatenatingnwiththemthpowerofm.
Form=3,thefirstfewnumbersofthesequenceare11,28,327,464,5125,6216,7343,8512,9729,101000,111331,121728,.
.
.
132Anaturalquestiontoaskistodeterminehowmanyprimesareinthesequenceforspecificvaluesofm.
Thefollowingtheoremanswersthegeneralquestionforanyvalueofm.
Theorem:TheonlyprimethatoccursintheSmarandachennmsequenceforanyvalueofmis11.
Proof:Itisclearthatthegeneralterma(n)oftheSmarandachennmgeneralizedsequenceisgivenbya(n)=n*10k+nmwherek=d(nm),thenumberofdigitsofnm.
Itfollowsthata(n)=n{10k+nm-1).
Forn=1,a(n)=11(independentofm.
)andisaprime.
Ifn>1,obviouslya(n)isdivisiblebynandthereforeiscomposite.
Definition:TheReducedSmarandachennmsequenceisthesetofnumbersb(n)=a(n)/n,wherea(n)istheelementoftheSmarandachennmgeneralizedsequence.
Inthiscase,wehaveb(n)=10k+nm-1wherek=d(nm)isthenumberofdigitsofnm.
Form=3,wehavea(n):11,28,327,464,5125,6216,7343,8512,9729,101000,111331,121728,.
.
.
b(n):11,14,109,116,1025,1036,1049,1064,1081,10100,10121,.
.
.
Openproblem:HowmanytermsinthissequenceareprimeOpenproblem:Howmanytermsinthegeneralb(n)sequenceareprimeSection28SomeIdeasOntheSmarandachenknSequenceDefinition:Letk>0beafixedinteger.
TheelementsoftheSmarandachenknGeneralizedSequenceareformedbyconcatenatingnandk*n.
Thenthtermofthesequenceisgivenbya(n)=k*n+n*10r,wherer=d(k*n),thenumberofdigitsofk*n.
Example:Fork=2,thefirstfewtermsofthesequenceare12,24,36,48,510,612,714,816,.
.
.
1224,1326,1428,.
.
whereeachtermisformedbyconcatenatingnand2n.
Thenthtermisgivenbya(n)=2*n+n*10r,wherer=d(2n),thenumberofdigitsof2n.
Ithasbeenconjecturedthatthenumberofperfectsquaresintheprevioussequence133(fork=2)isfinite.
Inthischapter,weanalyzethesequencefork=8andprovethatthereareinfinitelymanyperfectsquaresinthesequence.
Fork=8,thefirsttermsofthesequenceare18,216,324,432,540,648,756,864,972,1080,1188,1296,13104,14112,…Thegeneraltermisgivenbya(n)=8*n+n*10r,wherer=d(8*n),thenumberofdigitsof8*n.
Proposition:a(n)isaperfectsquareifn=(10k+8)/9.
Proof:Letn=(10k+8)/9,whichhasexactlykdigitsanda(n)isgivenbya(n)=8*{10k+8}/9+[{10k+8}/9]*10ka(n)={(10k+8)/9}*{8+10k}a(n)=9*{(10k+8)/9}*{(8+10k)/9}a(n)=9*{(10k+8)/9}2a(n)=(3n)2=aperfectsquare.
Thiscompletestheproof.
Thesequenceofnumbersdefinedbytheformula(10k+8)/9is12,112,1112,11112,andthecorrespondingn8nsequenceis1296,112896,11128896,1111288896,.
.
.
Whichisthesameas362,3362,33362,333362.
.
.
Openproblem:Otherthanthosedefinedbythisformula,howmanyperfectsquaresareinthesequenceAnexampleis324=182,whichdoesnotmatchthepattern.
Section29SomeNotionsOnLeastCommonMultiplesDefinition:TheSmarandacheLCMsequence(SLS)isdefinedbyTn=leastcommonmultipleofallintegers1throughn.
134Thefirstfewnumbersofthissequenceare1,2,6,60,60,420,840,2520,2520,.
.
.
Itiswellknownthatn!
dividestheproductofanysetofnconsecutivenumbers.
WewillusethisideaincombinationwiththeSLStodefineanothersequence.
Definition:ThetermsoftheSmarandacheLCMRatiosequenceoftherthkind(SLRS(r))aregivenbyrTn=LCM(n,n+1,n+2,.
.
.
n+r-1)/LCM(1,2,3,4,.
.
.
r)Examples:SLRS(1)1,2,3,4,5,1Tn(=n).
SLRS(2)1,3,6,10,.
.
.
2Tn=n(n+1)/2(triangularnumbers).
SLRS(3)LCM(1,2,3)/LCM(1,2,3),LCM(2,3,4)/LCM(1,2,3),LCM(3,4,5)/LCM(1,2,3),LCM(4,5,6)/LCM(1,2,3),LCM(5,6,7)/LCM(1,2,3)or1,2,10,10,35.
.
.
SLRS(4)1,5,5,35,70,42,210,.
.
.
Note:Itappearsthatforr>2,thesequencesdonotfollowapattern.
Openproblem:SearchforpatternsintheSLRS(r)sequencesandfindreductionformulasfortheelementsrTn.
Definition:Forn≥rnLr=LCM(n,n-1,n-2,.
.
.
n-r+1)/LCM(1,2,3,.
.
.
r)wherethenumeratoristheleastcommonmultipleofthelastrnumbersuptonandthedenominatoristheleastcommonmultipleofthefirstrnumber.
Bydefinition,wewillhave0L0=1.
Startingatthebeginning,wehave1350L0=11L0=1,1L1=1,2L0=1,2L1=2,2L2=2.
Whichcanbeusedtoformthetriangle11,11,2,11,3,3,11,4,6,2,11,5,10,10,5,11,6,15,10,5,1,11,7,21,35,35,7,7,11,8,28,28,70,14,14,2,11,9,36,84,42,42,42,6,3,11,10,45,60,210,42,42,6,3,1,1Definition:ThetriangleformedfromthenLrnumbersiscalledtheSmarandacheAMARLCMtriangle.
Note:Asr!
dividestheproductofrconsecutiveintegerssodoestheLCM(1,2,3,…r)dividetheLCMofanyrconsecutivenumbersTherefore,theelementsoftheSmarandacheAMARLCMTriangleareallintegers.
ThefollowingpropertiesoftheSmarandacheAMARLCMTriangleareeasytosee.
1.
Thefirstcolumnandtheleadingdiagonalelementsareallunity.
2.
ThekthcolumnaretheelementsofSLRS(k).
3.
ThefirstfourrowsarethesameasthatofthePascal'sTriangle.
4.
Thesecondcolumnisthesetofnaturalnumbers.
5.
Thethirdcolumnisthesetofthetriangularnumbers.
6.
Ifpisaprimethenpdividesallthetermsofthepthrowexceptthefirstandthelastwhichareunity.
InotherwordsΣpthrow≡2(modp).
Bycarefulobservation,additionalproblemspresentthemselves.
Forexample,intheninthrow,42appearsinthreeconsecutiveplaces.
Openproblem:DosequencesofequalvaluesofarbitrarylengthappearintheSmarandacheAMARLCMtriangle136Openproblem:Findaformulaforthesumoftherows.
Openproblem:Searchforcongruencepropertieswhenniscomposite.
TheSmarandachefunctionS(n)=k,isdefinedasthesmallestintegersuchthatndividesk!
.
Thefollowingfunctionisdefinedinasimilarway.
Definition:TheSmarandacheLCMfunctionSL(n)=k,isdefinedasthesmallestintegerksuchthatndividesLCM(1,2,3,k).
Letn=p1a1p2a2p3a3.
.
.
prarbetheprimefactorizationofnandletpmambethelargestdivisorofnwithonlyoneprimefactor,thenitfollowsthatSL(n)=pmam.
Ifn=k!
thenS(n)=kandSL(n)>k.
IfnisaprimethenwehaveSL(n)=S(n)=n.
ClearlySL(n)≥S(n)theequalityholdingfornaprimeorn=4,n=12.
AlsoSL(n)=nifnisaprimepower.
(n=pa).
Openproblem:Aretherenumbersn>12forwhichSL(n)=S(n)Openproblem:AretherenumbersnforwhichSL(n)=S(n)≤nSection30AnApplicationoftheSmarandacheLCMSequenceandtheLargestNumberDivisibleByAlltheIntegersNotExceedingItsrthRootDefinition:Thenumberswhicharedivisiblebyallnumbersnotexceedingtheirsquarerootare2,4,6,8,12and24.
ThesenumberswillbecalledtheSmarandacheMurtynumbersoforder2.
ThelargestnumberwillbecalledtheSmarandachePatinumberoforder2.
Thenumberswhicharedivisiblebyallwholenumbersnotexceedingtheircuberootare2,3,4,5,6,7,allevennumbersfrom8to26,36,48,60,72,…120,180,240,300,420.
Wehavesixconsecutivenumbersfrom2to7,tenconsecutiveevennumbersfrom8to26,eightconsecutivemultiplesof12from36to120,fourconsecutivemultiplesof60from120to300andfinally420,whichistheSmarandachePatinumberoforder3.
Ifweaddtheseconsecutivepropertiestogether,wehave6+10+8+4+1=29.
Note:As73=343729=93andfurthermore2520>1331=113,itisevidentthat420isthelargestnumberwhichisdivisiblebyallthewholenumbersnotexceedingitscuberoot.
Letm=[n1/3],where[]standsforthegreatestintegerfunction.
ThennwillbeaSmarandacheMurthynumberoforder3iftheLCMofthenumbersfrom1tomdividesn.
137OurnextpointofconsiderationwillbetofindanupperlimitfortheSmarandachePatinumberoforderr.
Considerthecaseforr=3.
Obviously,nisdivisiblebyalltheprimeslessthanm.
Lettherebesprimeslessthanm.
L(m)isalsodivisiblebyalltheprimesp1,p2,.
.
.
psandletL(m)=p1k1*p2k2p3k3.
.
.
psks.
Also,bythechoiceofthenumberswehavetheinequalitypiki≤n1/3n1/3orncomplete.
Definition:TheSmarandacheReducedMultiplesequenceisformedbydividingthetermsofthenknsequencebyn.
Thefirstfewtermsofthesequenceare1,12,123,120304,102030405,10203040506,1020304050607,.
.
.
a(13)=13263952657891104117130143156169b(13)=1020304050607008009010011012013.
Openproblem:HowmanytermsoftheSmarandachereducedmultiplesequenceareprimeSection32MoreontheSmarandacheSquareandHigherPowerBasesDefinition:TheSmarandacheSquareBaseistheexpressionofanumberasthesumofdistinctsquaresgreaterthanonepluse,wheree=0,1,2,or3.
ItisknownthateverypositivenumbercanbeexpressedinSmarandachesquarebaseform.
Inasimilarmanner,theSmarandachebasecubeandhigherpowerbasescanbedefined.
Definition:TheSmarandacheSquarePartResidueZerosequenceisthesetofnumbersthatcanbeexpressedasthesumoftwoormoreperfectsquaresthataregreaterthanorequaltoone.
Thefirstfewtermsofthissequenceare5,10,13,14,17,20,21,25,26,29,30,34,.
.
.
Openproblem:HowmanyofthenumbersintheSmarandachesquarepartresiduezerosequenceareperfectsquares139Note1:Allnumbersoftheform(a2+b2)2{thelargestmembersofthePythagoreantriplets}aremembersofthesequence.
DoallthesquaresinthissequencehavethisformNote2:Fornote1,itfollowsthatinfinitelymanyfourthpowersaremembersofthesequence.
Openproblem:HowmanyfifthorhigherpowersaremembersofthesequenceOpenproblem:HowmanyelementsinthesequencecanbeexpressedasthesumofsquaresintwoormorewaysSomeexamplesare45=36+9=25+16+4,125=100+25=121+4.
Openproblem:CanonefindasequenceofconsecutiveintegersofarbitrarylengthinthesequenceAlongsimilarlinestheSmarandacheSquarepartresidueunitytheSmarandacheSquarepartresiduetwoandtheSmarandacheSquarepartresiduethreesequencescanalsobedefined.
ThesameideascanbeusedtodefinetheSmarandcheCubePartResidueZerosequence.
Thefirstfewtermsinthesequenceare9,28,35,36,65,72,73,91,92,100,.
.
.
Openproblem:HowmanyelementsintheSmarandchecubepartresiduezerosequenceareperfectcubesSomeexamplesare216,3375,9261.
216=13+23+33=1+8+27,3375=153=13+23+33+43+63+113+123=1+8+27+64+216+1331+1728,213=9261=13+33+63+93+153+173=23+63+133+143+643.
Openproblem:HowmanyelementsinthesequencecanbeexpressedasthesumofperfectcubesinmorethanonewaySomeexamplesare:1729(theRamanujannumber)=123+1=103+93,213=9261=13+33+63+93+153+173=23+63+133+143+643.
Openproblem:IsitpossibletofindasequenceofconsecutiveintegersorarbitrarylengthintheabovesequenceThesquareofeverytriangularnumber{n*(n+1)/2}2,n>1isamemberoftheabovesequence.
Also,sincethereareinfinitelymanysquaretriangularnumbers,theabovesequencecontainsinfinitelymanyfourthpowers.
140Openproblem:HowmanyfifthorhigherpowersareinthesequenceSection33SmarandacheFourthandHigherPatterned/AdditivePerfectPowerSequencesConsiderthefollowingpatternedfourthpowersequence994,9994,99994,999994,.
.
.
wherethefourthpowersare96059601,996005996001,9996000599960001,99996000059999600001,.
.
.
Proposition1:Thenumber(10n–1)4isamemberofSmarandachefourthpoweradditivesequenceifn=24m+3*34k+2-1,forallmandk.
Proof:Itcanbeprovedthatthesumofthedigitsd(Tn)ofthenthtermTnisgivenbyd(Tn)=9+6+5+9+6+1+18*(n-1)=18*(n+1).
=2*32*(n+1).
Ifn+1=24m+3*34k+2thend(Tn)=24m+4*34k+4={2m+1*3k+1}4whichisaperfectfourthpower.
Sincetherearenorestrictionsonthevaluesofmandk,thisgeneratesinfinitelymanytermsoftheSmarandachefourthpoweradditivesequence.
Note:Moregenerally,ncanbechosenasr4*24m+3*34k+2–1.
(r,m,kchosenarbitrarily.
)TheSmarandachePatterned/Additivefifthpowersequence.
995,9995,99995,999995,.
.
.
9509900499,995009990004999,99950009999000049999,.
.
.
Inthissequence,thesumofthedigitsd(Tn)ofthenthtermTnaregivenbyTn=54+27(n-1)=27(n+1).
Andifn=r5*35m+2-1,theconditionsaresatisfied.
SmarandachePatternedsixthpowersequence.
996,9996,99996,999996,.
.
.
941480149401,994014980014994001,999400149980001499940001,.
.
.
Itcanbeprovedthatthereisnoterminthissequenceforwhichthesumofthedigitsisaperfectsixthpower.
Openproblem:ArethereinfinitelymanytermsintheSmarandachesixthpoweradditive141sequenceSmarandachePatterned/Additiveseventhpowersequence.
997,9997,99997,999997,.
.
.
93206534790699,993020965034979006999,9993002099650034997900069999,.
.
.
d(Tn)=72+36*(n-1)=36(n+1).
Forn=r7*97k+3–1wesatisfytheconditionsforwhichd(Tn)={r*9(k+1)}7.
SmarandachePatterned/Additiveeighthpowersequence.
998,9998,99998,999998,.
.
.
9227446944279201,992027944069944027992001,99920027994400699944002799920001d(Tn)=72+36(n-1)=36(n+1),Forn=r8*98k+5–1wesatisfytheconditionsforwhichd(Tn)={r*9(k+1)}8.
Additionalproblemtoconsider:1)Determinethevaluesofmforwhichthesequence(10n–1)mgivesinfinitelymanytermsoftheSmarandachemthpowerAdditivesequence.
(Form=6thereisnosuchterm).
Conjecture:ForeverymthereareinfinitelymanytermsintheSmarandachemthpoweradditivesequence.
Section34TheSmarandacheMultiplicativeCubicSequenceandMoreIdeasonDigitSumsDefinition:TheSmarandachemultiplicativecubesequenceisdefinedasasequenceofperfectcubesinwhichtheproductofthedigitsisalsoacube.
Thefirstfewtermsofthesequenceare:1,8,24389,226981,9393931,11239424,17373979,36264691,66923416,94818816,.
.
.
Whichare1,23,293,613,2113,2243,2593,3313,4063,4563,.
.
.
respectively.
Itisanopenproblemastowhetherthissequenceisfiniteorinfinite.
Toprepareforanattackonthisproblem,twoadditionalproblemswillbeexaminedfirst.
Problem1:Arethereinfinitelymanycubeswherenodigitiszero142Problem2:IsthereanypatternofnumbersthatgenerateperfectcubeswithnozerodigitItseemsthattheanswertoboththesetwoquestionsisno.
Someadditionalsequencesbasedonthedigitsumwillnowbegiven.
Smarandachesequenceofnumberswherethedigitssumtoaperfectsquare.
Thefirstfewtermsare1,4,9,10,13,18,22,27,31,36,40,45,5479,81,90,97,100,103,108,112,117,121,126,130,135,144,153,162,169,171,178,180,187,196,202,207,211,216,220,225,.
.
.
SmarandachesequenceofthesmallestnumberTnwhosedigitssumton2,d(Tn)=n2,Tnisthesmallestsuchnumber.
Thefirstfewtermsofthesequenceare:1,4,9,79,799,9999,499999,19999999,999999999,199999999999,.
.
.
TheSmarandachesequenceofthesmallestsquaresTnwhosedigitssumton2d(Tn)=n2,whered(Tn)=thesumofthedigitsofthenthterm,Tnisthesmallestsuchnumber.
Thefirstfewtermsofthissequenceare:1,4,9,169,4489,69696,.
.
.
Openproblem:Findaformulaforthegeneraltermofthissequence.
TheSmarandachesequenceofnumberswithdigitswhosefirstdigitsumisaperfectcube.
Thefirstfewtermsofthesequenceare:1,8,10,17,26,35,44,53,62,71,80,100,107,116,125,134,.
.
.
800,999,1000,1007,1016,.
.
.
1899,1989,1998,2006,2015,.
.
.
2799,2979,2997,3005,.
.
.
19999999,.
.
.
TheSmarandachesequenceofthesmallestnumberswhosedigitssumton3.
Thefirstfewtermsinthissequenceare:1,8,999,19999999,…TheSmarandachesequenceofthesmallestcubeswhosedigitssumton3.
Thefirstfewtermsofthesequenceare:1,8,19683,999400119992,999998500000749999875,.
.
.
1,23,273,99983,999953,9999999999993,.
.
.
d(999400119992)=64,d(999998500000749999875)=125d{(999999999999)3}=d(999999999997000000000002999999999999)=216Onasimilarline,thegeneralquestionsoftheSmarandachesequenceofthesmallestnumberwhosedigitssumtonm,andtheSmarandachesequenceofthesmallestmth143powerwhosedigitssumtonmcanbeposed.
Openproblem:Forwhatvaluesofm>3,istheSmarandachesequenceofthesmallestmthpowerwhosedigitssumtoanmthpower,finiteSection35SmarandachePrimeGeneratorSequenceIthasbeenproventhatforeveryprimep,thereisaprimeoftheformkp+1.
Thisisevidentfromthefactthatallthedivisorsofthenumber(10p-1)/9areoftheformk*p+1,whichitselfisoftheformk*p+1.
Startingfrom2,formasequenceofprimesinwhichT1=2,Tn+1=k*Tn+1,wherekisthesmallestnumberyieldingaprime.
Thefirstfewtermsofthissequenceare2,3,7,29,59,709,2837,.
.
.
Thesmallestprimenotincludedinthissequenceis5.
Then,startingwithT1=5,Tn+1=k*Tn+1,wherekisagainthesmallestnumberyieldingaprime,wegetthesequence,5,11,23,47,283,1699,.
.
.
Thesmallestprimenotalreadyinasequenceis13,sostartingwith13andthenrepeatingtheprocessusingadditionalprimes,wegetthefollowingsequences13,53,107,643,7717,.
.
.
17,103,619,2477,.
.
.
19,191,383,769,7691,.
.
.
31,311,1867,.
.
.
TheSmarandachePrimeGeneratorsequencewillbeconstructedusingthefirsttermsofthesesequences.
2,5,13,17,19,31,.
.
.
Conjecture:TheSmarandacheprimegeneratorsequenceisfinite.
IsthereanyprimethatisamemberofmorethanoneofthesequencesthatgeneratetheSmarandacheprimegeneratorsequenceLetpbeacommonmemberoftwosequenceswiththefirsttermasp1andp2respectively.
Then,forsomek1andk2wegetP=2k1*p1+1=2k2*p2+1,ork1*p1=k2*p2,ork1=r*p2andk2=r*p1.
Conjecture:AllofthesequencesthatgeneratetheSmarandacheprimegenerator144sequencearedistinct.
AdditionalopenproblemsthatIthoughtofregardingtheSmarandacheadditiveprimesequencewillclosethischapter.
Openproblem:AretherearbitrarilylongsequencesofconsecutiveprimeshavingthesamefirstsumofdigitsOpenproblem:AretherearbitrarilylongsequencesofconsecutiveprimeshavingthesamefinalsumofdigitsOpenproblem:AretherearbitrarilylongsequencesofconsecutivetermsoftheSmarandacheadditiveprimesequencewiththesamefirstsumofdigitsOpenproblem:AretherearbitrarilylongsequencesofconsecutivetermsoftheSmarandacheadditiveprimesequencewiththesamefinalsumofdigitsSection36ChapterReferences1.
AmarnathMurthy,"OpenProblemsandConjecturesontheFactor/reciprocalPartitionTheory",SNJVol.
11,No.
1-2-3,(2000).
308-311.
2.
AmarnathMurthy,"GeneralizationofPartitionFunction.
IntroducingSmarandacheFactorPartition",SNJVol.
11,No.
1-2-3,(2000).
227-239.
3.
MaohuaLe,"OntheBaluNumbers",SNJVol.
12,No.
1-2-3,(2001).
331-334.
4.
AmarnathMurthy,"SmarandachePascalDerivedSequences",SNJ,2000.
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AmarnathMurthy,"MoreResultsandApplicationsoftheSmarandacheStarFunction",SNJ,Vol.
11,No.
1-2-3,2000.
6.
Problem2/31,Mathematics&InformaticsQuarterly,3/99Volume9,Sept'99,Bulgaria.
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AmarnathMurthy,"PropertiesoftheSmarandacheStarTriangle",SNJ,Vol.
11,No.
1-2-3,2000.
8.
V.
Krishnamurthy,CombinatoricsTheoryandApplications,EastWestPressPrivateLimited,1985.
9.
AmarnathMurthy,"MiscellaneousresultsandtheoremsonSmarandacheFactorPartitions.
",SNJ,Vol.
11,No.
1-2-3,2000.
14510.
FlorentinSmarandache,Definitions,solvedandUnsolvedProblems,Conjectures,andTheoremsinNumberTheoryandGeometry,editedbyM.
L.
Perez,XiquanPublishingHouse,2000.
11.
Dr.
FanelIacobescu,"SmarandachePartitionTypeandOtherSequences",BulletinofPureandAppliedSciences,Vol.
16E(No.
2)1997,page240.
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AmarnathMurthy,"OnTheInfinitudeofPrimeNumbers",TheMathematicsEducation,India,Vol.
XXIX,No.
1,March,1995.
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FeliceRusso,ASetofNewSmarandacheFunctions,SequencesandConjecturesinNumberTheory,AmericanResearchPress,Lupton,AZ.
14.
AmarnathMurthy,"ExploringSomeNewIdeasonSmarandacheTypeSets,FunctionsandSequences",SNJ,Vol.
11,No.
1-2-3,Spring,2000.
15.
HenryIbstedt,"TheSmarandacheSequenceInventory",SmarandacheNotionsJournal,Vol.
10,1-2-3,Spring,1999.
16.
MaohuaLe,"TheReducedSmarandacheSquare-DigitalSubsequenceisInfinite",SNJ,Vol12,No.
1-2-3,Spring,2001.
17.
AmarnathMurthy,"SmarandachePythagorasAdditiveSquareSequence"(toappear).
18.
AmarnathMurthy,"FabricatingPerfectSquaresWithaGivenValidDigitsSum",(toappear).
19.
AmarnathMurthy,"OntheDivisorsofSmarandacheUnarySequence",SmarandacheNotionsJournal,Vol11.
,No.
1-2-3,Spring,2000.
146Chapter3MiscellaneousTopicsInthepreviouschapters,anunderlyingsimilarityorthemewasusedtogroupthematerialtogether.
However,notalloftheSmarandachenotionscanbesoeasilycategorized.
Therefore,thischapterservesasarepositoryforthosetopicsthatareconsidereddifferentenoughthatplacingtheminoneofthepreviouschaptersisconsideredinappropriate.
Section1ExploringSomeNewIdeasOnSmarandacheTypeSets,FunctionsAndSequences(1)SmarandachePatternedPerfectSquareSequences.
Considerthefollowingsequenceofnumbers13,133,1333,13333,1)whichisformedbythesquaresofthenumbers169,17689,1776889,177768889,2)Sequence(1)iscalledtherootsequenceof(2)anditisclearthatthereisapatterntothenumbersofbothsequences.
Therootsequenceisaonefollowedbyasequenceofn3'sandtheelementsoftheproductsequenceareaone,followedby(n-1)sevens,a6,(n-1)eightsandendinginanine.
Thereareafinitenumberofsuchpatternedperfectsquaresequencesandhereisalistoftherootsequences.
(I)13,133,1333,13333,.
.
.
(2)16,166,1666,16666,.
.
.
(3)19,199,1999,19999,.
.
.
(4)23,233,2333,23333,.
.
.
(5)26,266,2666,26666,.
.
.
(6)29,299,2999,29999,.
.
.
Alongsimilarlines,wehaverootsequenceswherethefirsttermsare(7)33(8)36(9)39(10)43(11)46(12)49(13)53(14)66(15)73(16)79(17)93(18)96(19)99.
OpenProblems:(1)Arethereanypatternedperfectcubesequences147(2)Arethereanypatternedperfectpowersequencesforapowergreaterthan3SmarandacheBreakupSquareSequencesThesequence4,9,284,61209,.
.
.
isdefinedby4=2249=7249284=22224928461209=702032TnisthesmallestsequenceofdigitssuchthattheconcatenationT1T2.
.
.
Tn-1Tnisaperfectsquare.
Thefollowinglimit(T1T2.
.
.
Tn-1Tn)1/2limitn→∞10kwherekisthenumberofdigitsinthenumerator,isclosetoeither2.
22.
.
.
or7.
0203.
.
.
SmarandacheBreakupCubeSequencesAlongsimilarlines,theSmarandacheBreakupCubeSequence,wherecubesareusedinsteadofsquares.
Byusinglargerexponents,wecandefineSmarandacheBreakupPerfectPowerSequences.
SmarandacheBreakupIncrementedPerfectPowerSequences1,6,6375,…1=11,16=42,166375=553,etc.
TnisthesmallestnumberwhosedigitsconcatenatedwiththepreviousnumbersinthesequenceT1T2T3.
.
.
Tnyieldsaperfectnthpower.
148SmarandacheBreakupPrimeSequenceTheSmarandacheBreakupPrimeSequenceisformedbyfindingthesmallestnumberTnsuchthatifitisconcatenatedontherighttotheconcatenationofallpreviousterms,aprimenumberisformed.
Inotherwords,2,3,3,.
.
.
2,23,233etc.
areprimes.
T1T2.
.
.
Tn-1TnisaprimeSmarandacheSymmetricPerfectSquareSequenceThissequenceisthesetofperfectsquaresthatarealsopalindromic.
Thefirstfewtermsare1,4,9,121,484,14641,.
.
.
SmarandacheSymmetricPerfectCubeSequenceThissequenceisthesetofperfectcubesthatarealsopalindromic.
Thefirstfewtermsare1,8,343,1331,.
.
.
SmarandacheSymmetricPerfectPowerSequenceThisisthegeneralsequence,forn≥2,thesetofnpowersthatarealsopalindromes.
SmarandacheDivisiblebynSequenceThetermsofthissequence(Tk)arethesmallestnumberssuchthatkdividesT1T2.
.
.
Tk.
Thefirstfewtermsofthissequenceare1,2,0,4,0,2,.
.
.
Whichisaconsequenceofthecomputations1|1,2|12,3|120,4|1204,5|12040,6|120402,.
.
.
SmarandacheSequenceofNumbersWheretheSumoftheDigitsIsPrime2,3,5,7,11,12,14,16,20,21,23,25,29,.
.
.
149SmarandacheSequenceofPrimesWheretheSumoftheDigitsIsPrime2,3,5,7,11,23,29,41,43,47,61,67,83,89,.
.
.
SmarandacheSequenceofPrimespwhere2p+1isAlsoPrime2,3,5,11,23,29,41,53,.
.
.
SmarandacheSequenceofPrimespwhere2p-1isAlsoPrime3,7,19,31,.
.
.
SmarandacheSequenceofPrimespwherep2+2isAlsoPrime3,17,.
.
.
SmarandacheSequenceoftheSmallestPrimesWhichDiffersby2nFromthePreviousPrime5,17,29,97,.
.
.
S1=5=3+2,S2=17=13+4,S3=29=23+6,S4=97=89+8.
SmarandacheSequenceofSmallestPrimeForWhichp+2risPrimeElementristhesmallestprimep,suchthatp+2risprime.
Thefirstfewtermsare3,13,23,89,.
.
.
Since3+2*1=5,13+2*2=17,23+2*3==29,89+2*4=97.
SmarandacheSequencesnoftheSmallestNumberWhoseSumofDigitsisnThefirstfewelementsare1,2,3,4,5,6,7,8,9,19,29,39,49,59,69,79,89,99,199,299,399,499,599,.
.
.
ThisisasequenceofnumberssatisfyingthepropertykN+1=Π(ar+1)r=1wherearistherthdigitofthenumber.
Proof:LetN=arar-1.
.
.
a1suchthat150kN+1=Π(ar+1)(3)r=1Thelargestk-digitnumberisN=10k-1,whereallthedigitsare9.
Itcanbeverifiedthatthisisasolution.
ArethereothersolutionsLetthemthdigitbechangedfrom9toam(amcomes10(k-1)(am+1).
Thisamountstothereductioninvalueby10(k-1)(9-am).
Thevalueofthek-digitnumberNgoesdownby10(m-1)(9-am).
Forthenewnumbertobeasolutionthesetwovalueshavetobeequalwhichoccursonlyatm=k.
Thisgives8moresolutions.
Inallthereare9solutionsgivenbya.
10k-1,fora=1to9.
Fork=3thesolutionsare199,299,399,499,599,699,799,899,999,…Question:ArethereinfinitelymanyprimesinthissequenceSmarandacheSequenceofNumbersSuchThattheSumoftheDigitsDividesn1,3,6,9,10,12,18,20,21,24,27,30,36,40,42,45,48,50,54,60,63,72,80,81,84,90,100,102,108,110,112,114,120,126,132,133,135,140,144,150,.
.
.
SmarandacheSequenceOfNumbersSuchThatEachDigitDividesn1,2,3,4,5,6,7,8,9,10,11,12,15,20,22,24,30,33,36,40,44,50,55,60,66,.
.
.
SmarandachePowerStackSequenceForn(SPSS(n))SPSS(2)Thenthtermisobtainedbyconcatentatingthedigitsofthepowersof2startingfrom20to2nandmovinglefttoright.
Thefirstfewdigitsofthissequenceare1,12,124,1248,12416,1241632,.
.
.
SPSS(3)Thenthtermisobtainedbyconcatentatingthedigitsofthepowersof3startingfrom30to3nandmovinglefttoright.
Thefirstfewdigitsofthissequenceare1,13,139,13927,.
.
.
151Question:Ifnisanoddnumbernotdivisibleby5,howmanytermsintheSPSS(3)sequenceareprimeItisclearthatn|snifandonlyifn≡0(mod5).
SmarandacheSelfPowerStackSequence(SPSS)Thekthterminthesequenceisformedbyconcatentatingthenumbers11,22,33kkstartingfromtheleft.
Thefirstfewtermsare1,14,1427,1427256,14272563125,142725631257776,.
.
.
SmarandachePerfectSquareCountPartitionSequence(SPSCPS(n))Thekthterm(startingatzero)ofthissequenceisdefinedasthenumberofperfectsquaresmthatsatisfytheinequalitynk+1≤m≤nk+n.
Forn=12,thefirstfewtermsinthesequenceSPSCPS(12)are3,1,2,.
.
.
sincethenumberofperfectsquareslessthan12is3andthenumberofperfectsquaresbetween13and24is1.
SmarandachePerfectPowerCountPartitionSequence(SPPCPS(n,k))Therthterm(startingatzero)ofSPPCPS(n,k)isthenumberofkthpowersmthatsatisfytheinequalitynr+1≤m≤nr+n.
Forexample,thefirsttermofSPPCPS(100,3)is4,as13,23,33,43arealllessthan100.
Question:DoesΣ(Tr/(nr))convergeasn→∞SmarandacheBertrandPrimeSequenceAccordingtoBertrand'spostulate,thereexistsaprimebetweennand2n.
Startingfrom2,formasequencebytakingthelargestprimelessthandoublethepreviousprimeinthesequence.
Thefirstfewelementsofthesequenceare2,3,5,7,13,23,43,83,163,.
.
.
152SmarandacheSemi-perfectNumberSequenceAsemi-perfectnumberisonethatcanbeexpressedasthesumofasubsetofitsdistinctdivisors.
Forexample,12=2+4+6=1+2+3+620=1+4+5+1030=2+3+10+15=5+10+15=1+3+5+6+15.
Itisclearthateveryperfectnumberisalsosemi-perfect.
Theorem:Thereareinfinitelymanysemi-perfectnumbers.
Proof:WeshallprovethatN=2np,wherepisprimelessthan2n+1–1isasemi-perfectnumber.
ThedivisorsofNare1,2,222np,2p,22p,2np.
Summingthesecondrow,wehaven-1∑2rp=p(1+2+22+23+.
.
.
2n-1)=p(2n-1)=M.
r=0ThedifferencebetweenNandMisp.
Itisknownthateverynumberisexpressibleasthesumofpowersoftwoandwehaveselectedpsothatitislessthanthelargestpowerintheabovelist.
Therefore,wecanexpresspasthesumofpowersoftwonp=∑ar*2r,wherear=0orar=1.
r=0Sincenoneofthesefactorswereusedintheprevioussum,wecanexpressNasthesumofasubsetofits'divisors.
Remark:Therearemanyadditionalexamplesofsemi-perfectnumbers.
Readersareencouragedtosearchforadditionalfamiliesofsemi-perfectnumbers.
SmarandacheCo-primeButNoPrimeSequence153Thissequencestartswithfourandeachsubsequenttermisthesmallestnumberthatisrelativelyprimetotheprevioustermandisnotprime.
Moreformally,ThenthtermTnisdefinedasfollowsTn={x|(Tn-1,x)=1,xisnotaprimeand(Tn-1,y)≠1forTn-1(pr-1+pr+1)/2.
Forexample:3(67+73)/2isastrongprime.
SmarandacheWeakPrimeSequence3,7,13,19,23,29,31,37,.
.
.
SmarandacheStrongPrimeSequence11,17,41,.
.
.
SmarandacheBalancedPrimeSequence5,.
.
.
154Itisclearthatforabalancedprime>5,pr=pr-1+6k.
Section2FabricatingPerfectSquaresWithaGivenValidDigitSumIntroduction:WhilestudyingtheSmarandacheadditivesquaresequence[48-1](sequenceofsquaresinwhichthedigitssumisalsoasquare),aquestionpoppedintomymind.
GivenanumberdcanonegetaperfectsquarewhosedigitsumisdInthischaptersomeresultspertainingtothisquestionhavebeenestablished.
Definition:Givenanyintegeranan-1.
.
.
a0thedigitsumisan+an-1+.
.
.
a0.
Ifthissumhasmorethanonedigit,repeatedlytakethesumofthedigitsuntiltheresultisaone-digitnumber.
Definition:Anumberdiscalledavaliddigitssumifd≡1(mod3),ord≡0(mod9)1)PropositionI:Thedigitsumofaperfectsquareisavaliddigitsum.
Proof:Considerthesquaresofnumbers1through91,4,9,16,25,36,49,64,81thedigitsumsare1,4,9,7,7,9,4,1,9whichdefinitelyareoftype(1).
Itcanalsobeprovedusingthepropertiesofcongruencethatthedigitssumoftheproductoftwonumbersistheproductofthedigitssums.
Sincethiscanberepeatedanarbitrarynumberoftimes,theproofiscomplete.
Theorem:Ifdisavaliddigitsumthenthereexistinfinitelymanyperfectsquareswhosedigitssumisd.
Proof:ConsiderthefollowingfourSmarandachePatternedperfectsquaresequences[48-2]alongwiththeirrootsequences.
155(I)9,99,999,9999,.
.
.
81,9801,998001,99980001,.
.
.
WehaveTn=10n-1,.
Tn2=102n-2*10n+1.
=10n+1(10n-1-1)+8*10n+1.
Hence,thesumofthedigitsofTn2=9(n-1)+8+1=9n.
(II)1,19,199,1999,.
.
.
1,361,39601,3996001,.
.
.
Tn=2*10n-1-1,Tn2=4*102(n-1)-4*10n-1+1.
Tn2=3*102(n-1)+10n*(10n-2-1)+6*10n-1+1.
ThesumofthedigitsofTn2=3+9(n-2)+6+1=9(n-1)+1.
(III)2,29,299,2999,.
.
.
4,841,89401,8994001,.
.
.
Tn=3*10n-1-1,Tn2=9*102(n-1)-6*10n-1+1.
Tn2=8*102(n-1)+10n*(10n-2-1)+4*10n-1+1.
ThesumofthedigitsofTn2=8+9(n-2)+4+1=9(n-1)+4.
(IV)5,59,599,5999.
.
.
25,3481,358801,35988001,.
.
.
Forn=1andn=2,thesumofthedigitsofTn2are7and16respectively.
Forn≥3Tn=6*10n-1-1,andTn2=36*102(n-1)-12*10n-1+1Tn2=35*102(n-1)+10n+1*(10n-3-1)+88*10n-1+1.
ThesumofthedigitsofTn2=3+5+9(n-3)+8+8+1=9(n-1)+7.
Thispatternalsoholdsforn=1and2aswell.
Sinceeverynumberofthetype3*r+1iscongruentto1,4or7(mod9)thepreviousfourcasescoverallthevaliddigitssums.
Therefore,wehaveprovedthatthereexistsaperfectsquarewithagivenvaliddigitssum.
Byaddinganevennumberofzeroswegetinfinitelymanysuchnumbersandtheproofofthetheoremiscomplete.
156Example:Letd=124,soourgoalistofindaperfectsquareNwithdigitssum=124.
Wehaved=124≡7(mod9).
Therequirednumberisthememberofsequence(IV),forwhichn=14,[124=9(14-1)+7.
]N=(6*1013-1)2=3599999999999880000000000001.
Conjecture:Foragivenvaliddigitssumdthereexistsinfinitelymanynontrivialperfectsquareswhosedigitssumisd.
(IfNisasolutionthenN*102nisanontrivialone.
)Generalizingthisresult,ifwedefineavaliddigitsumforacubeasanumbercongruentto0,1,or8(mod9),thenwecanputforwardthefollowingconjectures.
Conjecture:Foragivenvaliddigitsumdthereexistsinfinitelymanynontrivialperfectcubeswhosedigitssumisd.
(IfNisasolutionthenN*103nisanontrivialone.
)Conjecture:Foragivenvaliddigitsumdthereexistsinfinitelymanynontrivialperfectmthpowerswhosedigitsumisd.
(IfNisasolutionthenN*10m*nisanontrivialone.
)Section3FabricatingPerfectCubesWithaGivenValidDigitSumIntroduction:Intheprevioussection,givenanarbitrarynumberd,thequestiontoconsiderwaswhethertherewasaperfectsquarewithadigitsumequaltod.
Inthissection,wewillconsiderthesimilarproblemwheresquareisreplacedbycube.
Definition:Anumberdiscalledavaliddigitsumforacubeisd≡0,1,or8(mod9).
Inotherwords,thedigitsumis9k,9k+1or9k–1.
Theorem:Foragivenvaliddigitsumforcubed,thereexistsinfinitelymanyperfectcubeswhosedigitsumisd,whendisoftheform18k,9k+1or9k-1.
Toprovethistheorem,westartwiththefollowingproposition.
Proposition:Foraperfectcube,thedigitsumnecessarilyisavaliddigitssumforacubesatisfyingcondition(1).
Proofoftheproposition:Examiningthecubesofthenumbers1through9,weget1,8,27,64,125,216,343,512,and729.
Thecorrespondingdigitssumsare1,8,9,10,8,9,10,8,18,allofwhichreducedownto1,8or9.
Thisisconsistentwiththedefinitionofthevaliddigitssumforacube.
Also,itcanbeprovedusingthepropertiesofcongruencethatthedigitssumoftheproductoftwonumbersistheproductofthedigitssums.
Theproofiscomplete.
157Proofofthetheorem:ConsiderthefollowingSmarandachepatternedperfectcubesequenceswiththecorrespondingrootsequences(I)991,9991,99991,999991,.
.
.
973242271,997302429271,999730024299271,999973000242999271wehaveTn=10n+2-9,fortherootsequence,andTn3=103n+6-27*102n+4+243*10n+2-729.
ThegeneraltermofthecubesequenceTn3=tnisgivenbytn=102n+7*(10n-1-1)+973*102n+4+242*10n+2+103*(10n-1-1)+271.
Uponsimplificationwehavetn=103n+6-27*102n+4+243*10n+2-729=Tn3.
Thesumofthedigitsoftn=9(n-1)+9+7+3+2+4+2+9(n-1)+2+7+1equals18(n+1)+1=9(2m)+1,m>1.
With253=15625,thesumofthedigitsis19,thecasewherem=1isalsoincluded.
(II)995,9995,99995,999995,.
.
.
985074875,998500749875,999850007499875,999985000074999875.
WehavefortherootsequenceTn=10n+2-5,Tn3=103n+6-15*102n+4+75*10n+2-125.
Forthecubesequencetn=Tn3tn=102n+7*(10n-1–1)+985*102n+4+74*10n+2+103*(10n-1–1)+875.
Whenthisissimplified,wehavetn=103n+6–15*102n+4+75*10n+2–125=Tn3.
158Thesumofthedigitsoftnequals9(n-1)+9+8+5+7+4+9(n-1)+8+7+5=18(n-1)+53=9(2m)-1,m>2.
With173=4913,thesumofthedigitsis17,and953=857375,sumofdigits=35,thecasem=1andm=2arealsoincluded.
(III)9,99,999,9999,.
.
.
729,970299,997002999,999700029999,.
.
.
ConsidertherootsequenceTn=10n-1,Tn3=103n-3*102n+3*10n-1.
Forthecubesequencetn=Tn3tn=102n+1*(10n-1-1)+7*102n+2*10n+(10n-1).
Whenthisexpressionissimplified,wehavetn=103n-3*102n+3*10n-1=Tn3.
Thesumofthedigitsforthisnumberequals9(n-1)+7+2+9n=18n=9(2n)=9(2m).
Withtheabovethreesequenceswehavetakencareofthedigitsums9k,9k+1and9k-1,forkeven.
Wewillnowconsiderthesequencesforkodd.
(IV)97,997,9997,99997,.
.
.
912673,991026973,999100269973,999910002699973,.
.
.
ConsidertherootsequenceTn=10n+1–3,Tn3=103n+3–9*102n+2+27*10n+1–27.
Forthecubesequencetn=Tn3tn=102n+4(10n-1–1)+91*102n+2+26*10n+1+100*(10n-1–1)+73.
WhichyieldsthefollowingwhensimplifiedTn=103n+3–9*102n+2+27*10n+1–27=Tn3.
159Thesumofthedigitsfortn=9(n-1)+9+1+2+6+9(n-1)+7+3=18(n-1)+28=9(2n+1)+1.
With73=343,digitssum=10,whichisthecasewheren=0.
(V)98,998,9998,99998,.
.
.
941192,994011992,999400119992,999940001199992,.
.
.
WehavetherootsequenceTn=10n+1–2,Tn3=103n+3–6*102n+2+12*10n+1–8.
Forthecubesequencetn=Tn3tn=102n+4(10n-1–1)+94*102n+2+11*10n+1+100*(10n-1–1)+92.
WhichonsimplificationgivesTn=.
103n+3–6*102n+2+12*10n+1–8=Tn3.
Thesumofthedigitsfortn=9(n-1)+9+4+1+1+9(n-1)+9+2=18(n-1)+26=9(2n+1)-1.
83=512,withdigitsum8,isthecasewheren=0.
Therefore,thetheoremisprovenford=18k,9k+1and9k-1.
Example:Givend=118,findaperfectcubeNsuchthatthedigitsumisequaltod.
Wehaved=118=9*13+1.
HenceNisamemberofsequence(IV)andN=(1014-3)3.
Note:Readersareencouragedtolookforotherexhaustivesetsofsequences.
OpenProblem:Findasequenceofcubes,thesumofwhosedigitsisanoddmultipleof9.
Considerthefollowingtable160NN3SumofdigitsofN3=dd/932791333593727333336926037364333337025927037455333333703592593703763733333337036925926037037728333333337037025925927037037819333333333703703592592593703703799113333333333703703692592592603703703710812Basedontheabovetable,wemakethefollowingconjectures:ConjectureI:(a)IfN=(103k–1)/3thenthesumofthedigitsofN3is9(4k).
ConjectureI:(b)IfN=(103k-1–1)/3thenthesumofthedigitsofN3is9(4k–1).
ConjectureI:(b)IfN=(103k+1–1)/3thenthesumofthedigitsofN3is9(4k+1).
ConjectureII:Foragivenvaliddigitsumdthereexistinfinitelymanynontrivialperfectcubeswhosedigitsumisd.
(IfNisasolutionthenN*103nisanontrivialone.
)Note:IfconjectureIistrue,itwilltakecareofd=9(2k+1),anoddmultipleof9.
TogetherwiththetheoremI,itwouldleadtothetruthofconjectureII.
Section4SmarandachePerfectPowersWithGivenValidDigitSumIn[2]theSmarandacheadditivesquaresequenceisdefinedasthesequenceofsquaresinwhichthedigitsumisalsoasquare.
Thevaliddigitsumforasquarewasdefinedinaprevioussectionasanumberdsuchthatd≡1(mod3)ord≡0(mod9).
Inthissection,wedefineaSmarandachesequenceofperfectsquareswithagivendigitfirstsumasthesequenceofperfectsquareswhosedigitssumisthesame.
Examples:Ifdigitsum=1,thereisthesequence1611,100,10000,102n,.
.
.
Fordigitsum=41,4,121,10201,.
.
.
Fordigitsum=716,25,1024,2401,.
.
.
Fordigitsum=99,36,81,144,225,324,441,900,.
.
.
Fordigitsum1064,361,.
.
.
Fordigitsum1349,256,625,841,.
.
.
Usingthefirsttermsoftheabovesequences,wecandefinetheSmarandachesequenceofsmallestperfectsquareswithvaliddigitfirstsums1,4,16,9,64,49,169,576,289,.
.
.
wherethedigitsumsare1,4,7,9,10,13,16,18,19,.
.
.
Openproblem:Therearethreeconsecutivetermsinincreasingorder49,169,576.
IsitpossibletohaveanarbitrarynumberoftermsinincreasingordecreasingorderWedefinetheSmarandachesequenceofperfectsquareswithagivendigitsfinalsum{1}1,64,100,289,361,676,784,1225,1369,.
.
.
wheretherootsequenceis1,8,10,17,19,26,28,35,37,a).
Additionalsequencesare{4}4,49,121,256,400,625b)(7)16,25,169,196,484,529,c)162{9}9,36,81,144,225d).
NotethatwehaveTn=9*n2insequence(d).
OpenProblem:Tofindanexpressionforthenthtermforthesequences(a),(b),and(c).
Theaboveideacanbegeneralizedbydefining(A)Smarandachesequenceofperfectcubeswithagivendigitfirstsumaslistedbelowwherethenumberinthebraces{}isthedigitsum.
{1}1,1000,1000000,.
.
.
103n.
.
.
{8}8,125,512,1331,.
.
.
{9}27,216,27000,.
.
.
{10}64,343,64000,.
.
.
{17}2744,4913,12167,.
.
.
{18}729,1728,3375,5832,9261,13824,91125,.
.
.
Wehavesimilarsequencesfor19,26,27,28andsoforth.
Notethatforallthenumbersd,d≡0,1or8(mod9).
(B)Smarandachesequenceofperfectcubeswithagivendigitfinalsum.
{1}1,64,343,1000,4096,6859,.
.
.
{8}8,125,512,1331,2744,4913,12167,.
.
.
{9}27,216,729,Tn=27*n3.
OpenProblem:Tofindthegeneraltermforthesequenceswherethesumsare1and8.
(C)Smarandachesequenceofsmallestperfectcubeswithvaliddigitfirstsumsisdefinedinthefollowingway:1,8,27,64,2744,729,2197,.
.
.
withthedigitsums1,8,9,10,17,18,19,.
.
.
Openproblem:Wehavefiveconsecutivetermsinincreasingorder:1,8,27,64,2744.
CanwehaveanarbitrarynumberoftermsinincreasingordecreasingorderGeneralization:TheSmarandacheSequenceofperfectmthpowerswithagivendigitfirstsum,theSmarandacheSequenceofperfectmthpowerswithagivendigitfinalsumandtheSmarandachesequenceofsmallestperfectmthpowerswithvaliddigitfirstsumscanallbedefinedonsimilarlines.
163Section5NumbersThatAreaMultipleoftheProductofTheirDigitsAndRelatedIdeasSmarandacheProudPairsofNumbersDefinition:Wesaythatapairofintegers(m,n)isaSmarandacheproudpairifn=m=Pd(n),wherePd(n)istheproductofthedigitsofn.
Examples:Forthesingledigitnumbersn=1*n36=2*(3*6),15=3*(1*5),24=3*(2*4),175=5*(1*7*5)So,(1,1),(2,2),(3,3)9,9),(3,15),(3,24),(5,175)areSmarandacheproudpairs.
Conjecture:Foreverymhavingnozerodigit,thereexistsanumbernsuchthat(m,n)isaSmaradacheproudpair.
Conjecture:Foreverym,thereexistsinfinitelymanynsuchthat(m,n)isaSmarandacheproudpair.
NumbersForWhichthemthPoweroftheSumoftheDigitsEqualstheSumoftheDigitsofthemthPowerNumbersforWhichtheSquareoftheSumoftheDigitsEqualstheSumoftheDigitsoftheSquareIfd(n)isthesumofthedigitsofn,thenthesenumberssatisfytheformulad(n2)=[d(n)]2.
Examples:112=121,(1+1)2=1+2+1122=144,212=441,(2+1)2=4+4+1222=484,(2+2)2=4+8+4.
132=169,312=961,(1+3)2=1+6+91112=12321,(1+1+1)2=1+2+3+2+12122=44944,(2+1+2)2=4+4+9+4+4.
164Thereareinfinitelymanysuchnumbersisevidentfromthefactthat{10k+1)and2{10k+1)satisfytheconditionsforallvaluesofk.
Ifthesearetobeconsideredabittrivial,thenhereareadditional,nontrivialpatterns.
n=212,2102,21002,210002,.
.
.
n2=44944,4418404,441084004,.
.
.
n=122,1022,10022,100022,.
.
.
n2=14884,1044484,100440484,.
.
.
NumbersForWhichtheCubeoftheSumoftheDigitsEqualstheSumOftheDigitsoftheCubeInthiscase,wearelookingforsolutionstod(n3)=[d(n)]3Thefamily101,1001,10001,canbeconsideredtrivialexamples.
Nontrivialexamplesare113=1331,(1+1)3=1+3+3+11113=1367631,(1+1+1)3=1+3+6+7+6+3+1.
Numbersoftheform1011,10011,100011,.
.
.
and1101,11001,110001,.
.
alsosatisfytheexpression.
Thetwoadditionalexamplesare:(1010010001)3=1030331606363361603330030001d(1030331606363361603330030001)=64(11010010001)3=1334636937969963961633330030001d(1334636937969963961633330030001)=125whichsuggeststhefollowingconjecture.
Conjecture:Foreverypositiveintegermthereexistsanumbernsuchthatm3=d(n3)={d(n)}3.
NumbersForWhichtheFourthPoweroftheSumoftheDigitsEqualstheSumoftheDigitsoftheFourthPowerInthiscase,wearesearchingforsolutionstotheexpressiond(n4)=[d(n)]4.
Examples:114=14641,(1+1)4=1+4+6+4+1101,1001,10001,.
.
.
165Openproblem1:Aretherenumbersn,suchthatm4=d(n4)=[d(n)]4,wherem>2Openproblem2:Aretherenumbersn,suchthatd(nk)=[d(n)]k,wherek>4Generalization:AdditionalSmarandachedigitalsequencescanbedefinedbystudyingrelationsbetweenafunctionofthedigitsandanotherfunctionofthenumberitself.
mthpowerswherepermutationsofthedigitsarealsomthpowersForm=2144=122,441=212,169=132,196=142,961=312,1296=362,2916=542,9216=962,9261=213,1089=332,9801=992.
1024=322=210,2401=492=74,4761=692,1764=422,1936=442,1369=3721782=31684,1912=36481,1962=38416=144,2092=43681.
Form=3125=53,512=83,3313=36264691,4063=66923416.
Form=4256=44,625=54.
Openproblem:Aretherenumbersmandnsuchthatthedigitsofmkareapermutationofthedigitsofnkforallk>1Section6TheLargestandSmallestmthPowerWhoseDigitSum/ProductIsIts'mthRootIntroduction:WhilestudyingtheSmarandacheadditivesquaresequence[7](sequenceofsquaresinwhichthedigitsalsosumtoasquare),aproblemoccurredtome.
ArethereperfectsquareswhosedigitsumisthesameasthesquarerootIfso,thentheremustbeasmallestandalargestsuchnumber.
Asimilarquestioncanbeaskedforhigherpowers.
Inthissectionthatquestionisexamined.
WewillrefertonumbersthatareperfectpowersofthesumoftheirdigitsSmarandacheAnuragnumbers.
Thatsuchnumbersexistcanbeseenfrom92=81,andthesumofthedigitsof81=9.
Forpurposesofthisanalysis,wewillconsideronetobeatrivialsolutionandignoreit.
Wewillshowthat81istheonlynontrivialperfectsquarewiththispropertyandwewillcallittheSmarandacheShikhanumberfortwo.
Bydefault,81isthelargestperfectsquarewhosedigitssumtothesquareroot.
Itisalsothesmallestsuchnumber,soitwillalsobetheSmarandacheAnirudhnumberfor2.
166Wewillnowprovethat81istheonlySmarandacheAnuragnumber.
Asthereisnoperfectsquarenumberwiththispropertysmallerthan1296=362,and36beingthelargestpossiblesumofa4digitnumberitisevidentthat81istheonlysuchperfectsquare.
Movingontothecubes,wehave83=512=(5+1+2)3,173=4913=(4+9+1+3)3,183=5832=(5+8+3+2)3,263=17576=(1+7+5+7+6)3and273=19683=(1+9+6+8+3)3.
As453=91125(afivedigitnumber)and45isthelargestpossiblesumofa5-digitnumber,itisevidentthat19683isthelargestcubewiththispropertyandistheSmarandacheShikhanumberfor3.
Letthesmallestsuchnumber(8)becalledtheSmarandacheAnirudhnumberfor3.
Consideringthefourthpowerswehave74=2401=(2+4+1)4,224=234256,254=390625,284=614656,and364=1679616.
Thereisnonumberbetween364and724with724=26873856(aneightdigitnumber).
Since72isthelargestpossiblesumofan8-digitnumber,364=1679616istheSmarandacheShikhanumberfor4and7istheSmarandacheAnirudhnumberfor4.
Consideringthefifthpowerswehave285=17210368,355=52521875,365=60466176,and465=205962976.
Itistobenotedthatonlynumberswithfinalsumofdigits1,8or9qualifytobeSmarandacheAnuragnumbersforfive.
Conjecture:Foreverym>2,thereexistsatleasttwoSmarandacheAnuragnumbers.
Inotherwords,theSmarandacheShikhaandSmarandacheAnirudhnumbersaredistinct.
Conjecture:ThetotalnumberofSmarandacheAnuragnumbersforthemthpowersaremorethatthatforthe(m+1)thpower.
Definition:Anumber(wherenodigitiszero)divisiblebytheproductofits'digitsiscalledaSmarandacheMeenakshiNumber.
ThesequenceobtainedbyapplyingthispropertyiscalledtheSmarandacheMeenakshiSequence.
Thefirstfewtermsare1,2,9,12,15,24,36,.
.
.
LetPd(n)denotetheproductofthedigitsofN.
Withthisnotation,wehave167Pd(36)=18,Pd(144)=92916=542.
Pd(2916)=2*9*1*6=108,2916/108=27248832=125,Pd(248832)=3072=248832/3072=81429981696=1446,Pd(429981696)=1679616,429981696/1679616=256.
Proposition:ThereareinfinitelymanytermsintheSmarandacheMeenakshiSequence.
Proof:LetN=(10n–1)/9.
ItisclearthatPd(N)=1*1.
.
.
*1=1andisanelementoftheSmarandacheMeenakshiSequence.
Thesenumberswillbeconsideredtrivialsolutions.
Definition:Thefollowingwillbeconsideredsemi-trivialelementsoftheSmarandacheMeenakshiSequence.
12,112,1112,11112,.
.
.
15,115,1115,11115,.
.
.
Additionalsequencesare1113,1111113,1111111113,.
.
.
whereTn=10*(103n–1)/9+3,1111117,1111111111117,1111111111111111117,.
.
.
whereTn=10*(106n–1)/9+7.
Aproofthat7divideseachelementofthelastsequenceisgivenin[7].
AllothertermsoftheSmarandacheMeenakshiSequencewillbeconsiderednon-trivial.
Conjecture:Thereareinfinitelymanynon-trivialtermsintheSmarandacheMeenakshiSequence.
Openproblem:Isthereanymthpowerwhosemthrootequalstheproductofits'digitsInotherwords,aretheresolutionstotheequation[Pd(N)]m=NNote:IfNisasolutionto[Pd(N)]m=NthenitisevidentthatNtakesthecanonicalform,N=2a*3b*5c*7d,wherea,b,c,darenonnegativeintegers.
168Section7AConjectureond(N),theDivisorFunctionItselfAsADivisorwithRequiredJustificationIntroduction:ThenumberofdivisorsofanaturalnumbervariesquiteirregularlyasNincreases.
Itisknownthatd(N)2.
)Theorem:IfIistheindexofbeautyofMandifI=n1*n2*.
.
.
nristheSmarandachefactorpartition(abreakupofIastheproductofitsdivisors),thenJ=p1(n1-1)*p2(n2-1)*pr(nr-1)istheindexofbeautyofM*Jwhen(M,J)=1.
Proof:LetN=M*J(1),thend(N)=d(M)*d(J)as(M,J)=1.
d(N)=d(M)*n1*n2*.
.
.
nrd(N)=d(M)*Id(N)=M(2),asIistheindexofbeautyforM.
From(1)and(2)wegetN/d(N)=J.
Therefore,JistheindexofbeautyofM*J.
Corollary:IfIistheindexofbeautyofMthenpI-1istheindexofbeautyforM*pI-1if(M,p)=1.
Thefollowingresultsprovidethemotivationforthisconjecture.
169Definition:Forconvenienceweusethesymbolp[r]forp1*p2*p3*p4,theproductoffourprimes.
Whenexaminingthecontentsofthefollowingtable,thislistofpointsshouldbekeptinmind.
a)Allp'sandq'sareprimes.
b)IfN=M*prthen(M,p)=1.
c)IfN=M*p[r],then(M,p[r])=1.
S.
N.
INd(N)1p224p2,18p2,9p224,18,92p336p3363p440p4,60p440,604p572p5725p684p6846p796p7,80p796,807p8108p81088p9180p91809p10132p1013210p11240p1124011p[1]8p,12p8,1212p1p2=p[2]36p[2]3613p[3]96p[3]9614p[4]2532p[4]253215p[5]267p[5]26716p[7]295p[7]29517p[8]2113p[8]211318p[11]215p[11],212.
13p[11],2137p[11]215,212.
13,213.
719p[12]2143*5p[12]2143*520p[13]2145*32p[13]2145*3221p[16]2195p[16]219522p[20]22111p[20]2211123p[23]224523p[23]22452324p[p'-2]2p'-1*p'*p[p'-2]2p'-1*p'17025p[p2-2]2p2-1p23p[p2-2]2p2-1p2326p[p3-3]2p3–1*p3p[p3-3]2p3–1*p327p[p4-2]2p4–1p45p[p4-2]2p4–1p4528p[2p-3]22p-1*p*p[2p-3]22p-1*pIntherowsoftheprecedingtable,IistheindexofbeautyforthecorrespondingstructureofN.
Thoughtheconjecturecanbeestablished/verifiedforanumberofcanonicalforms,theproofforthegeneralcasewillmostlikelygivemathematiciansmanymoresleeplessnights.
Alongsimilarlines,thefollowingconjecturesareputforward.
Conjecture:ForeverypositiveintegerkthereexistsanumberNsuchthatN/S(N)=k,whereS(N)istheSmarandachefunction.
Examples:Fork=2,N=6,8;fork=3,N=12;fork=4,N=20;fork=5,N=50.
Conjecture:ForeverypositiveintegerkthereexistsanumberNsuchthatN/φ(N)=k.
(Euler'sfunction.
)Conjecture:Foreverypositiveintegerk,thereexistsanumberNsuchthatσ(N)/N=k.
(Fork=2wehaveN,aperfectnumber).
Section8SmarandacheFitorialandSupplementaryFitorialFunctionsTheSmarandacheFitorialFunction,denotedbyFI(N)isdefinedastheproductofalltheφ(N)numbersrelativelyprimetoandlessthanN.
Examples:FI(6)=1*5=5,FI(7)=6!
=720,FI(12)=1*5*7*11=385.
TheSmarandacheSupplementaryFitorialFunction,denotedbySFI(N)isdefinedastheproductofalltheremainingN-φ(N)numberslessthanorequaltoNwhicharenotrelativelyprimetoN.
Examples:SFI(6)=2*3*4*6=144,SFI(7)=7,SFI(11)=11,SFI(12)=2*3*4*6*8*9*10*12=1244160.
171Theorem:FI(N)andSFI(N)satisfythefollowingproperties:1.
FI(N)*SFI(N)=N!
.
2.
SFI(p)=p,andFI(p)=(p-1)!
iffpisaprime.
3.
FI(N)p.
3.
Thesumoftheφ(N)numbersrelativelyprimetoNisgivenby(Nφ(N))/2.
Therefore,thearithmeticmean(A.
M.
)ofthenumbersinthesumisN/2.
Theirgeometricmeanisgivenby1/φ(N)G.
M.
={FI(N)}Usingtherelationshipbetweenthesemeanswhenthenumbersarenotallequal(A.
M.
>G.
M.
),wehave1/φ(N){FI(N)}G2andfinally172SFI(N)Combining(4)and(5)weget(A).
Thefollowingresultisadirectconsequenceofformula(A).
5.
SFI(N)SFI(N)OpenProblem-III:Ifd(n)isthenumberofdivisors,forwhatvaluesofNisd(FI(N))>d(SFI(N))OpenProblem-IV:ForwhatvaluesofNisd(FI(N))σ(SFI(N))Definethesumoftheφ(N)relativelyprimenumbersasA(N)andthatoftheremainingN-φ(N),numbersasB(N).
ThenwehaveA(N)=Nφ(N)/2B(N)=N(N+1)/2-Nφ(N)/2.
Forexample,ifN=12,A(12)=24,andB(N)=54andforN=15,A(15)=60,andB(N)=60.
Note:A(N)=B(N),iffφ(N)=(N+1)/2,providedthatNisodd.
Readersareencouragedtoexplorethisfurther.
OpenProblem:ForwhatvaluesofN,isittruethatA(N))>B(N))ThecasewhereN=3nisconsideredbelow.
ForN=3nweget,A(N)=3n*{3n*(1-1/3)}/2=32n-1B(N)=3n*{3n+1}/2-32n-1={32n+3n-2*32n-1}/2={32n-1+3n}/2Forn=13n=32n-1,forn>1,2n-1>nand32n-1>3n,henceB(N)A(N).
Section9SomeMoreConjecturesOnPrimesandDivisorsThereareaninnumerablenumberofconjecturesandunsolvedproblemsinnumbertheorybasedonprimenumbers,whichhavebeengivingmathematicianssleeplessnightsallovertheworldforcenturies.
Hereareafewmoretoaddtotheirtroubles:(1)Everyevennumbercanbeexpressedasthedifferenceoftwoprimes.
(2)Everyevennumbercanbeexpressedasthedifferenceoftwoconsecutiveprimes.
i.
e.
foreverymthereexistsannsuchthat2m=pn+1–pn,wherepnisthenthprime.
174(3)EverynumbercanbeexpressedasN/d(N),forsomeN,whered(N)isthenumberofdivisorsofN.
Ifd(N)dividesN,wedefineN/d(N)=IastheindexofbeautyforN.
Conjecture:ForeverynaturalnumberMthereexistsanumberNsuchthatMistheindexofbeautyforN.
InotherwordsM=N/d(N).
Theconjectureistrueforprimes,whichiseasilyproven.
Wehave2=12/d(12)=12/6,2istheindexofbeautyfor12.
3=9/d(9)=9/3,3istheindexofbeautyfor9.
Foraprimep>3wehaveN=12p,d(N)=12andN/d(N)=p.
(N=8pcanalsobeused).
Theconjectureistrueforalargenumberoffamiliesofnumbers.
Howevertheproofofthegeneralcaseisstillunsolved.
(4)Ifpisaprime,thenthereareinfinitelymanyprimesoftheform(A)2np+1.
(B)2*anp+1.
(5)Itisawell-knownfactthatonecanhavearbitrarilylargenumbersofconsecutivecompositenumbers.
Forexampleforanyvalueofr:(r+1)!
+2,(r+1)!
+3,(r+1)!
+4,.
.
.
(r+1)!
+r-1,(r+1)!
+risalistofrconsecutivecompositenumbers.
Butthisisnotnecessarilythesmallestsetofsuchnumbers.
Letusconsiderthesmallestsetofrconsecutivecompositenumbersforthefirstfewvaluesofr.
rSmallestsetofcompositenumbersR/firstcompositenumber111/128,92/8314,15,163/14424,25,26,274/24524,25,26,27,285/24690,91,92,93,94,956/90790,91,92,93,94,95,967/908114,115,…1218/114175Similarlyfor9,10,11,12,13thefirstofthecompositenumbersis114.
Conjecture:Thesumoftheratiosinthethirdcolumnisfiniteand>e.
(6)GivenanumberN,carryoutthefollowingstepstogetanumberN1.
N–pr1=N1,wherepr1compositenumbersfromn+1throughn+m.
Sincerisnotprime,itmustbepossibletofactoritintotwofactors,eachofwhichis≥2.
Letr=p*qbethatfactorization.
Ifoneofthefactors(sayq),is≥n,thenr=p*q≥2n.
But,accordingtoBertrand'spostulate,theremustbeaprimebetweennand2n,whichyieldsacontradictionoftheassumptionthatallofthenumbersn+1throughn+marecomposite.
Therefore,eachofthefactorsmustbelessthann.
Therearetwopossibilities:176Case1:p≠q.
Inthiscase,eachfactorislessthannsop*q=rdividesn!
.
Case2:p=qwheretheyareprime.
Thismeansthatr=p2andtherearethreesubcases.
Subcase1:p=2.
Thenr=4andnmustbe3.
Since4doesnotdivide3!
,wehavethespecialcaseeliminatedinthestatementofthetheorem.
Subcase2:p=3.
Thenr=9andnmustbe7or8and9dividesbothofthesenumbers.
Subcase3:p≥5.
Thenr=p2>4p=>4p2pcompositenumbersbetweennand2n.
Note:ItiswellknownthatS(n)≤nforS(n)theSmarandachefunction.
Fromtheprevioustheorem,itispossibletodeducethefollowinginequality.
Ifpristherthprimeandpr≤n1+x,x>0,(1+1/n)(1+1/n)>1,n>0,178whichgives1/(r+1)n≥S2.
S3=S2+1/(r+k2)suchthatS3+1/(r+k2+1)>n≥S3.
Andsoon.
Then,thereexistsafinitemsuchthatSm+1+1/(r+km)=n.
Remarks:a)Thereareinfinitelymanydisjointsetsofnaturalnumbersthesumofwhosereciprocalsisunity.
b)Amongthesetsmentionedin(a),therearesetswhichcanbeorganizedinanordersuchthatthelargestelementofanysetissmallerthanthesmallestelementofthenextset.
Section12SmarandacheDeterminantSequencesDefinition:TheSmarandacheCyclicDeterminantNaturalSequenceisdefinedasthedeterminantsofthefollowingsequenceofmatrices1791121231234andsoon.
21231234131234124123Thedeterminantsofthefirstfourmatricesare1,-3,-18,and160.
TheseinitialvaluessuggestthefollowinggeneralformulaTn=(-1)[n/2]{(n+1)/2}*nn-1where[x]istheintegerpartofx.
Thisformulawillbeverifiedforthecasewheren=5,whichwilldemonstratehowthegeneralcaseishandled.
1234523451T5=345124512351234Bycarryingoutthefollowingelementaryrowoperationsa)R1=sumofalltherows.
b)Taking15fromthefirstrow.
c)ReplacingCk,thekthcolumnbyCk–C1,wehave10000123-12123-112-2-1=15312-2–1=151-3-2-141-3-2-1-4-3-2-15-4-3-2-1R1–R2,R3–R2,R4–R2,00501512-2-1=1875,thevaluesuggestedbythe0-500formula.
-5-500Althoughtheproofofthegeneralcaseisclumsy,itisbasedonsimilarlines.
180Generalization:Thissequencecanbefurthergeneralizedbyusinganarithmeticprogressionwithathefirsttermandcommondifferenced.
WewilldefinetheSmarandacheCyclicArithmeticdeterminantsequenceinthefollowingway:aaa+daa+da+2da+daa+da+2daandsoon.
a+2daa+dConjecture:Tn=(-1)[n/2]*Sn*dn-1*nn-2=(-1)[n/2]*{a+(n-1)d}*(1/2)*(nd)n-1whereSnisthesumofthefirstntermsofthearithmeticprogression.
Openproblem:Developaformulaforthesumofthefirstntermsofthissequence.
Definition:TheSmarandachebisymmetricdeterminantnaturalsequenceisthedeterminantsofthefollowingsequenceofmatrices.
112123123421232234332134324321wherethematricesaresymmetricacrossbothmaindiagonals.
Thedeterminantsofthesematricesare1,-2,-12,40,.
.
.
ThevaluesofthesefirstfewtermssuggeststhatthegeneralformulaisTn=(-1)[n/2](n(n+1))*2n-3Wewillverifythatthisformulaalsoholdsforn=5andthegeneralcasecanbedealtwithusingasimilarsequenceofoperations.
1234523454T5=345434543254321181Carryingoutthefollowingsequenceofrowoperationsa)R1=sumofalltherows.
b)Take15fromthefirstrowtoget123212101510-1-2-1-2-3-4R1=R1+R4gives000-212101510-1-2=120,whichisthevaluepredictedfromthesuggested-1-2-3-4generalformula.
Theproofofthegeneralcaseisbasedonsimilaroperations.
Generalization:Thissequenceofdeterminantscanbealsobegeneralizedusingtheelementsofanarithmeticprogression.
aaa+daa+da+2da+daa+da+2da+da+2da+daConjecture:ThegeneraltermofthissequenceofdeterminantsisgivenbyTn=(-1)[n/2]*(a+(n+1)d)*2n-3dn-1.
Section13ExpansionofxninSmarandacheTermsofPermutationsDefinition:Giventhefollowingexpansionofxnxn=b(n,1)x+b(n,2)x(x-1)b(n,n)xPnwedefineb(n,r)x(x-1)(x-2).
.
.
(x-r+1)(x-r)astherthSmarandachetermintheexpansion.
Inthissection,wewillexaminesomeofthepropertiesofthecoefficientsandencountersomefascinatingcoincidences.
Wewillstartbyexaminingthevaluesofsometermsforspecificvaluesofx.
182Forx=1,b(n,1)=1.
Forx=2,b(n,2)=(2n–2)/2.
Forx=3,b(n,3)=[3n–3–6(2n–2)/2]/6=(1/3!
)*(1*3n–3*2n+3*1n–1*0n).
Forx=4,b(n,4)=(1/4!
)*[1*4n–4*3n+6*2n–4*1n+1*0n].
Theseinitialvaluessuggestthefollowingtheorem.
Theorem:rb(n,r)=(1/r!
)∑(-1)r-k*rCk*kn=a(n,r).
k=1Firstproof:Thefirststepistoprovethepropositionb(n+1,r)=b(n,r-1)+r*b(n,r).
Startingwithxn=b(n,1)x+b(n,2)x(x-1)+b(n,3)x(x-1)(x-2)b(n,n)xPnreplacingxwithr,wehavern=b(n,1)r+b(n,2)r(r-1)+b(n,3)r(r-1)(r-2)b(n,n)rPn.
Multiplyingbothsidesbyrrn+1=b(n,1)r*r+b(n,2)r*r(r-1)+b(n,3)r*r(r-1)(r-2)b(n,r)r*rPr+termsequaltozero.
Usingslightlydifferentnotation,theexpressionisequivalenttorn+1=b(n,1)r*rP1+b(n,2)r*rP2+b(n,3)r*rP3+.
.
.
+b(n,r)r*rPr.
Usingtheidentityr*rPk+1+k*rPk,theexpressioncanberewrittenasrn+1=b(n,1){rP2+1*rP1}+b(n,2){rP3+2*rP2}b(n,r){rPr+r*rPr-1}183rn+1=b(n,1)rP1+{b(n,1)+2*b(n,2)}rP2+{b(n,2)+3*b(n,3)}rP3{b(n,r-1)+r*b(n,r)}rPrrn+1=b(n+1,1)rP1+b(n+1,2)*rP2+b(n+1,3)*rP3+.
.
.
+b(n+1,r)*rPr.
ThecoefficientsofrPt(tcomplete.
Secondproof:Thisproofisbasedonacombinatorialapproach.
Ifnobjects,wherenotwoarealike,aretobedistributedinxboxes,notwoalike,andeachboxcancontainanarbitrarynumberofobjects,thenumberofwaysthiscanbedoneisxn,sincetherearenalternativesfordisposalsofthefirstobject,nalternativesforthedisposalofthesecond,andsoon.
Alternately,letususeadifferentapproach.
Considerthenumberofdistributionsinwhichexactlynobjectsaretobeplacedinagivensetofrboxes(therestareempty).
Letthenumberofdistributionsberepresentedbyf(n,r).
Wederiveaformulaforf(n,r)byusingtheinclusion/exclusionprinciple.
Themethodisillustratedbythecomputationoff(n,5).
Considerthetotalnumberofarrangements,5nofnobjectsin5boxes.
Saythatsuchanarrangementhasproperty'a'.
Incasethefirstboxisempty,property'b'incasethesecondboxisempty,andsimilarproperty'c','d',and'e'fortheotherthreeboxesrespectively.
Tofindthenumberofdistributionswithnoboxempty,wesimplycountthenumberofdistributionshavingnoneoftheproperties'a','b','c',.
.
.
etc.
WecanapplytheformulaN-rC1.
N(a)+rC2.
N(a,b)-rC3.
N(a,b,c)+.
.
.
Here,N=5nisthetotalnumberofdistributions.
ByN(a),wemeanthenumberofdistributionswiththefirstboxempty,soN(a)=4n.
Similarly,N(a,b)isthenumberofdistributionswherethefirsttwoboxesareempty.
However,thisisthesameasthenumberofdistributionsinto3boxesandN(a,b)=3n.
Thus,wecanwriteN=5n,N(a)=4n,N(a,b)=3netc.
N(a,b,c,d,e)=0.
Applyingthepreviousformula,wegetf(n,5)=5n-5C1.
4n+5C2.
3n-5C3.
2n+5C4.
1n-5C5.
0n.
Bygeneralizingthisandreplacing5withr,wehavef(n,r)=rn-rC1*(r-1)n+rC2*(r-2)n-rC3*(r-3)n+.
.
.
rf(n,r)=∑(-1)krCk(r-k)nk=0185f(n,r)=r!
*a(n,r),fromtheorem(3.
1)ofref.
[12].
Now,theserboxesoutofthegivenxboxescanbechoseninxCrways.
Hence,thetotalnumberofwaysinwhichndistinctobjectscanbedistributedintoxdistinctboxes,occupyingexactlyrofthem(withx-rboxesempty),definedasd(n,r/x),isgivenbyd(n,r/x)=r!
*a(n,r)xCrd(n,r/x)=a9,wegetclosedsegmentsaspartofthefigure,withcompletetrianglesforn=10.
195A0B0A0B0A1x=2πx=πn=1n=2A1B1x=2π/3n=3A0B0B1A2A1π/2n=4A0B0B2A3A1A2B1A0B0π/3n=6196B2A2B1A1A0B02π/5n=5A1B1A0B0A2B2π/4n=8197A1B1A0B0A2B2B4A3A4B32π/9n=9A1A0B0B1A2A3B2B3B4A5A42π/10,n=10198A4B4B5A6B3A5A3A1π/6,n=12B2A0B0B1A2Section16SmarandacheRouteSequencesConsiderarectangularcitywithameshoftrackswhichareofequallengthandwhichareeitherhorizontalorverticalandmeetingatnodes.
Ifonerowcontainsmtracksandonecolumncontainsntracksthenthereare(m+1)(n+1)nodes.
Tobegin,letthecitybeofasquareshapei.
e.
m=n.
ConsiderthepossiblenumberofroutesRwhereapersonatoneendofthecitycantakefromasourceS(startingpoint)toreachthediagonallyoppositeendDthedestination.
199SA=(j,k-1),B=(j-1,k),C=(j,k)BACD(mrowsandmcolumns).
Refertothepreviousfiguretoseethederivationofthefollowingvalues.
Form=1,thenumberofroutesR=1.
Form=2,thenumberofroutesR=2.
Form=3,thenumberofroutesR=12.
Weseethatfortheshortestroutes,onehastotravel2munitsoftracklength.
Thereareroutesof2m+2unitsinlengthuptothelongestthatare4m+4unitslong.
Definition:TheSmarandacheRouteSequence(SRS)isdefinedasthenumberofallpossibleroutesforasquaremxmcity.
Therouteswillbeoflength2mupthrough4m+4.
Openproblem:DeriveageneralformulaforSRS.
Ournextstepherewillbetoderiveareductionformula,whichisageneralformulaforthenumberofshortestroutes.
Reductionformulaforthenumberofshortestroutes:LetRj,k=numberofshortestroutesfromStonode(j,k).
Node(j,k)canbereachedonlyfromnode(j-1,k)orfromnode(j,k-1),asonlyshortestroutesaretobeconsidered.
Itisclearthatthereisonlyonewayofreachingnode(j,k)fromnode(j-1,k).
Similarly,thereisonlyonewayofreachingnode(j,k)fromnode(j,k-1).
Hence,thenumberofshortestroutesto(j,k)isgivenbyRj,k=1*Rj-1,k+1*Rj,k-1=Rj-1,k+Rj,k-1.
200ThisgivesthereductionformulaforRj,k.
Applyingthisreductionformulatofillthechartweobservethatthetotalnumberofshortestroutestothedestination(theotherendofthediagonal)is2nCn.
Thiscanbeestablishedbyinduction.
Wecanfurthercategorizetheroutesbythenumberofturningpoints(TPs)itissubjectedto.
Thefollowingchartcontainsthenumberofturningpointsforacitywithninenodes.
No.
ofTPs1234No.
ofroutes2225Forfurtherinvestigation:1)Exploreforpatternsinthetotalnumberofroutes,numberofturningpointsanddevelopformulasforsquareaswellasrectangulargrids.
2)Tostudyhowmanyroutespassthroughagivennumber/setofnodes.
HowmanyofthempassthroughallthenodesSection17SmarandacheGeometricalPartitionsandSequences1)SmarandacheTraceableGeometricalPartitionConsiderachainhavingidenticallinks(sticks)whichcanbebentatthehingestogiveitdifferentshapes.
Forexample,considerthefollowingsetsofonethroughfourlinks.
(1)(2)(3)201(4)202Notethattheshapesofthefiguressatisfythefollowingrules:1)Thelinksareeitherhorizontalorvertical.
2)Nofigurecanbeobtainedfromanyotherbyrotationonly.
Itmustalsobeliftedfromthehorizontalplane.
3)Asthelinksareconnected,thereareonlytwoendsandonecantravelfromoneendtotheothertraversingallthelinks.
Thereareatthemosttwoends(therecanbezeroendsincaseofaclosedfigure)toeachfigure.
Thesearethenodeswhichareconnectedtoonlyonelink.
Definition:Fornthenumberofconnectedsticks,wedefinetheSmarandacheTraceableGeometricPartitionSgp(n)tobethenumberofdifferentfiguresthatcanbeconstructedusingthosehinged,connectedsticks.
ThesequenceofnumbersiscalledtheSmarandacheTraceableGeometricPartitionSequence(STGPS).
ThefirstfewnumbersofSTGPSare1,2,6,15,.
.
.
Openproblem:ToderiveareductionformulaforSTGPS.
203Definition:Abendisdefinedasapointwherethereisanangleof900betweentheconnectedsticks.
Thefollowingisatableofthenumberofsticksthathavespecificnumbersofbendsinthem.
NumberofbendsNo.
ofsticks01234110000211000312300413731Readersareencouragedtoextendthistableandlookforpatternsinthenumberofbends.
2)SmarandacheComprehensiveGeometricPartitionStartwithasetofidenticalsticksandconnectthemasbefore.
However,inthiscase,wewillrelaxthepreviousrulestoallow:1)Stickscanhavemorethanoneend.
2)Itmaynotbepossibletotravelfromoneendofthefiguretotheotherandtraverseeachstickonlyone.
Withthisrelaxationoftherulesdefiningthefigures,wehavethefollowingsetsoffigures.
(1)(2)204(3)(4)205206Definition:ThesequenceoffiguresthatcanbecreatedinthiswayiscalledtheSmarandacheComprehensiveGeometricPartitionFunction(SCGP)andthesequenceofnumbersformedbythenumberofsuchfiguresfornsticksisknownastheSCGPS.
207ThefirstfewtermsoftheSCGPSare1,2,7,25,.
.
.
Thefollowingtablesummarizesthenumberoffigureshavingaspecificnumberofends.
NumberofsticksNo.
ofends123400001100002126143001940001Thetablecanbeextendedbyincreasingthenumberofsticksandreadersareencouragedtosearchforpatternsinthetable.
Openproblem:DeriveareductionformulaforSCGPS.
Furtherconsideration:Thisideacanbeextendedbyallowingthebendstobeanglesotherthan900.
Section18SmarandacheLuckyMethodsinAlgebra,TrigonometryandCalculusDefinition:AnumberissaidtobeaSmarandacheLuckyNumberifanincorrectcalculationleadstoacorrectresult.
Forexample,inthefraction64/16ifthe6'sareincorrectlycanceled(simplified)theresult(4)isstillcorrect.
MoregenerallyaSmarandacheLuckyMethodissaidtobeanyincorrectmethodthatleadstoacorrectresult.
In[8],thefollowingquestionisasked:(1)ArethereinfinitelymanySmarandacheLuckyNumbersWeclaimthattheanswerisyes.
Alsointhissection,wegivesomefascinatingSmarandacheLuckymethodsinalgebra,trigonometryandcalculus.
ThefollowingaresomeexamplesofSmarandacheLuckyNumbers.
208(1)64/16=4/1=4.
(Cancelingthe6fromnumeratoranddenominator).
(2)95/19=5/1=5.
(Cancelingthe9fromnumeratoranddenominator).
(3)136/34=16/4=4.
(Cancelingthe3fromnumeratoranddenominator).
ThefollowingSmarandacheLuckyNumberscanbeusedtogeneratemanyadditionalluckynumbers,althoughthefamilycouldbeconsideredtrivial.
4064/1016,40064/10016,Ingeneral400000.
.
.
64/100000.
.
.
16inwhichboththenumeratoranddenominatorcontainthesamenumber(n)ofzeroesandthesixesarecancelled.
ASmarandacheLuckyMethodInTrigonometrySomestudentswhohavejustbeenintroducedtotheconceptoffunctionmisinterpretf(x)astheproductoffandx.
Inotherwords,theyconsidersin(x)tobetheproductofsinandx.
Thisgivesrisetoafunny,luckymethodapplicabletothefollowingidentity.
Toprovesin2(x)-sin2(y)=sin(x+y)sin(x-y)LHS=sin2(x)-sin2(y)={sin(x)+sin(y)}*{sin(x)-sin(y)}.
(A)Factoringthe"common"sinfromall"factors"={sin(x+y)}*{sin(x-y)}=RHS.
Thecorrectmethodfrompoint(A)onwardsshouldhavebeen{2sin((x+y)/2)*cos((x-y)/2)}*{2cos((x+y)/2)*sin((x-y)/2)}.
={2sin((x+y)/2)*cos((x+y)/2)}*{2cos((x-y)/2)*sin((x-y)/2)}.
={sin(x+y)}*{sin(x-y)}=RHS.
209ASmarandacheLuckyMethodInAlgebraInvectoralgebrathedotproductoftwovectors(a1i+a2j+a3k)and(b1i+b2j+b3k)isgivenby(a1i+a2j+a3k)(b1i+b2j+b3k)=a1b1+a2b2+a3b3Ifthissameideawasappliedtoordinaryalgebra(a+b)(c+d)=ac+bd.
B)Thiswrongluckymethodisapplicableinprovingthefollowingalgebraicidentity.
a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)RHS=(a+b+c)(a2+b2+c2-ab-bc-ca)=(a+b+c){(a2-bc)+(b2-ac)+(c2-ab)}applyingthewrongluckymethod(B),onegets=a.
(a2-bc)+b(b2-ac)+c(c2-ab)=a3-abc+b3-abc+c3-abc=a3+b3+c3-3abc=LHS.
ASmarandacheLuckyMethodInCalculusThefuninvolvedinthefollowingluckymethodincalculusistwofold,anditgoeslikethis.
Astudentisaskedtodifferentiatetheproductoftwofunctions.
Insteadofapplyingtheformulaforthedifferentiationoftheproductoftwofunctions,heappliesthemethodofintegrationoftheproductoftwofunctions(integrationbyparts)andgetsthecorrectanswer.
Theheightofcoincidenceisifanotherstudentisgiventhesameproductoftwofunctionsandaskedtointegratedoesthereverseofiti.
e.
heendsupapplyingtheformulafordifferentiationoftheproductoftwofunctionsandyetgetsthecorrectanswer.
IwouldtakethelibertytocallsuchaluckymethodtobeaSmarandachesuperluckymethod.
Considertheproductoftwofunctionsxandsin(x)f(x)=xandg(x)=sin(x).
d{f(x)g(x)}/dx=f(x)∫g(x)dx-∫[{d(f(x))/dx}∫g(x)dx]dxd{(x)sin(x)}/dx=(x)∫sin(x)dx-∫[{d(x)/dx}∫sin(x)dx]dx210=-(x)(cos(x))+sin(x)=-xcos(x)+sin(x).
TheSmarandacheluckymethodofintegrationisasfollows:∫{(f(x))g(x)}dx.
=f(x)d{g(x)}/dx+g(x)d{f(x)}/dx.
Ifweusethesamefunctionsusedinthepreviousexample,whenweapplythisluckymethodweget∫{(x)sin(x)}dx=(x){cos(x)}+{sin(x)}(1)or∫{(x)sin(x)}dx=xcos(x)+sin(x).
Byapplyingthecorrectmethods,itcanbeverifiedthatthesearethecorrectanswers.
References1.
DavidM.
Burton,"ElementaryNumberTheory,SecondEdition",pg.
171.
2.
FeliceRusso,"ASetofNewSmarandacheFunctions,SequencesandConjecturesinNumberTheory"pg.
64,AmericanResearchPress,LuptonUSA.
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AmarnathMurthy,"ExploringSomeNewIdeasonSmarandacheTypeSets,FunctionsandSequences",SmarandacheNotionsJournalVol.
11,No.
1-2-3,Spring2000.
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AmarnathMurthy,"GeneralizationOfPartitionFunction,IntroducingSmarandacheFactorPartition",SNJ,Vol.
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AmarnathMurthy,"Somemoreconjecturesonprimesanddivisors".
SNJVol.
12No.
1-2-3,2001.
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"TheFlorentineSmarandacheSpecialCollection",ArchivesofAmericanMathematics,CentreforAmericanHistory,UniversityofTexasatAustin,USA.
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AmarnathMurthy,"OnTheDivisorsOfSmarandacheUnarySequence",SNJ,vol.
11,No.
1-2-3,Spring2000.
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M.
L.
PerezandK.
Atatassov"Definitions,Conjectures,UnsolvedProblemsonSmarandacheNotions",AmericanResearchPress,1999.
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KrassimirT.
Atanassov,'"OnSomeOfTheSmarandache'sProblems"AmericanResearchPress,Lupton,AZUSA.
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(22-23).
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"TheFlorentineSmarandacheSpecialCollection",ArchivesofAmericanMathematics,CentreforAmericanHistory,UniversityofTexasatAustin,USA.
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"SmarandacheNotionsJournal"Vol.
10,No.
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AmarnathMurthy,"GeneralizationOfPartitionFunction,IntroducingSmarandacheFactorPartition",SNJ,.
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AmarnathMurthy,"AGeneralResultOnTheSmarandacheStarFunction",SNJ,.
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AmarnathMurthy,"MoreResultsAndApplicationsOfTheGeneralizedSmarandacheStarFunction",SNJ,.
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AmarnathMurthy,"ExpansionOfXnInSmarandacheTermsOfPermutations"SNJ,.
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212Indexφ-Sequence93Algorithm62-64ArithmeticMean172-173Atanassov,KrassimirT.
177BackwardExtendedFibonacciSequence(BEFS)82-83BalancedPrime154BaluNumbers30,72–75,BaluClass30BellNumbers24,45,47,77,99Bertrand'sPostulate152CProgrammingLanguage16,64CombinatorialSequences92Co-primePartitions(SPcp(n))92Depascalization81-83DigitSum155-157,157-161,161-163DiophantineEquation111DivisorMultipleSequence91DivisorPartitions92DivisorProductSequence94DivisorSequences90-91DullNumbers(*)30Euler(*)19Euler'sFunction(*)171ExtentOfANumber55-58Fermat'sLittleTheorem49FibonacciSequence77-78,82-83,97GeneralizedSmarandacheStarFunction40-46GeometricMean172-173Hardy19HarmonicSeries178213IndexofBeauty169-171,174-175Jacobi19Le,Maohua18,123L(N),LengthOfTheFactorPartition58-60MersennePrime88Non-Co-primePartitions(SPncp(n))92PairwiseCo-primeSets103-106PartitionSequences92-93PascalDerivedSequence78-79PascalSelfDerivedSequence77Pascal'sTriangle81Pascalization78,83-86PrimeDivisibilitySequence93-94PrimeLocationSequences91PrimePartition92PrimeProductSequences93PrimeSquarePartitions92PrincipleReciprocalPartition13PythagoreanTriplets139QuadPrimeSequenceGenerator91Ramanujannumber140ReciprocalPartitionTheory18ReducedSmarandachennmSequence132-133Smarandache(mth)PowerAdditivemthpowerSequence119Smarandache(mth)PowerAdditiventhpowerSequence119-120Smarandache(mth)PowerAdditiveSquareSequence119SmarandacheAccomodativeSequence(SAS)113SmarandacheAdditiveSquareSequence110-111SmarandacheAdditiveCubeSequence122-123SmarandacheAMARLCMTriangle136SmarandacheAnirudhNumbers167SmarandacheAnuragNumbers166-167214SmarandacheBalancedPrimeSequence154SmarandacheBeautifulNumber169SmarandacheBertrandPrimeSequence152SmarandacheBisymmetricDeterminantNaturalSequence181-182SmarandacheBreakupCubeSequences148SmarandacheBreakupIncrementedPerfectPowerSequences148SmarandacheBreakupPrimeSequence148-149SmarandacheBreakupSquareSequences148SmarandacheComprehensiveGeometricPartition203-206SmarandacheComprehensiveGeometricPartitionFunction(SCGP)206-207SmarandacheCyclicDeterminantNaturalSequence179-181SmarandacheDistinctReciprocalPartitionofUnitySet(SDRPS(n))11-18SmarandacheDistinctReciprocalpartitionofunitysequence12SmarandacheDivisorSequences60-61SmarandacheDivisorSumGeneratorSequence(SDSGS)116-117SmarandacheDivisorSum-divisorSumSequences(SDSDS).
116SmarandacheDivisorsofDivisorsSequence115SmarandacheDualSymmetricFunctions108-109SmarandacheFactorPartition(SFP)(*)19-21,28,46,50-54,55-58,58-60,62-64,64,72-75,77SmarandacheFactorPartitionSequence(*)29SmarandacheFactorialPrimeGenerator114SmarandacheFermatAdditiveCubicSequence128-130SmarandacheFibonacciBinarySequence(SFBS)96-97SmarandacheFibonacciProductSequence97-98SmarandacheFitorialFunctionFI(N)171-174SmarandacheForwardReverseSumSequence114SmarandacheFriendlyNumbers86-89SmarandacheFriendlyNaturalNumberPairs86SmarandacheFriendlyPair86-89SmarandacheFriendlyPrimePairs86SmarandacheFunctionS(n)136,171215Smarandache(Inferior)PrimePartSequence177SmarandacheIntegerPartenSequence88SmarandacheIntegerPartknSequences(SIPS)88SmarandacheIntegerPartπnSequence88SmarandacheLCMEvenSequence(SLES)90SmarandacheLCMfunctionSL(n)136SmarandacheLCMOddSequence(SLOS)90SmarandacheLCMRatioSequence134-135SmarandacheLCMSequence(SLS)89,134SmarandacheLuckyMethod208-210SmarandacheLuckyNumber207-208SmarandacheMaximumReciprocalRepresentationFunction(SMRR)178-179SmarandacheMaximumReciprocalRepresentationSequence(SMRRS)178-179SmarandacheMeenakshiNumbers167-168SmarandacheMultiplicativeCubeSequence142SmarandacheMultiplicativeSquareSequence111-112Smarandache-MurthyFigures192-198Smarandache-MurthyNumber137-138Smarandachen2nSequence126-128SmarandachenknGeneralizedSequence133-134,138-139Smarandachenn2Sequence130-132SmarandachennmSequence132-133SmarandacheOver-friendlyPair87SmarandachePascalDerivedSequences75-81SmarandachePascalDerivedBellSequence77SmarandachePascalDerivedCubeSequence79-80SmarandachePascalDerivedFactorialSequence80SmarandachePascalDerivedSquareSequence79SmarandachePascalDerivedTriangularNumberSequence80Smarandache-PatiNumber137-138SmarandachePatterned/AdditiveFifthPowerSequence141SmarandachePatternedPerfectSquareSequence110,147SmarandachePatternedPerfectCube122,147216SequenceSmarandachePatternedPowerSequence141-142SmarandachePerfectSquareCountPartitionSequence(SPSCPS(n))152SmarandachePowerStackSequenceForn(SPSS(n))151SmarandachePrimeGeneratorSequence143-144SmarandachePrime-PrimeSequence115SmarandacheProductofDigitsSequence88SmarandacheProudPairsofNumbers164SmarandachePythagorasAdditiveSquareSequence118-119SmarandacheReciprocalFunctionSc(n)176-177SmarandacheReducedDivisorSumPeriodicitySequences117SmarandacheReducedMultipleSequence138-139SmarandacheReverseMultipleSequence114SmarandacheRepeatableReciprocalpartitionofunity(SRRPS(n))11-16SmarandacheRepeatableReciprocalPartitionofUnitySequence11SmarandacheReverseAuto-CorrelatedSequence(SRACS)94-96SmarandacheReverseAutoCorrelationTransformation(SRACT).
94-95SmarandacheRouteSequence(SRS)199-200SmarandacheSelfPowerStackSequence(SPSS)151-152SmarandacheSemi-Accomodative113SmarandacheSemi-perfectNumberSequence152-153SmarandacheShikhaNumbers167SmarandacheSigmaDivisorPrimeSequence87SmarandacheSigmaPrimeSequence87SmarandacheSigmaProductOfDigitsEvenSequence89SmarandacheSigmaProductOfDigitsNaturalSequence88SmarandacheSigmaProductofDigitsOddSequence89SmarandacheSmallestNumberWith'n'DivisorsSequence87SmarandacheSquareBase139217SmarandacheSquare-DigitalSubsequence123SmarandacheSquarePartResidueZero139SmarandacheSquarePartResidueUnity139SmarandacheStarDerivedSequenceSd98-100SmarandacheStarFunctionF'*(N)22-23,30-39SmarandacheStarTriangle(SST)47-50,58,190-191SmarandacheStrictlyStairSequence100-101SmarandacheStrongPrimeSequence154SmarandacheSummableDivisorPairs(SSDP)88SmarandacheSupplementaryFitorialFunctionSFI(N)171-174SmarandacheSymmetricPerfectCubeSequence149SmarandacheSymmetricPerfectmthPowerSequence124-126SmarandacheSymmetricPerfectPowerSequence149SmarandacheSymmetricPerfectSquareSequence149SmarandacheTraceableGeometricalPartition200-202SmarandacheTraceableGeometricPartitionSequence(STGPS)202-203SmarandacheTriangular-TriangularNumberSequence115SmarandacheUnarySequence106-108SmarandacheUnder-FriendlyPair87SmarandacheVedamSequence29SmarandacheWeakPrimeSequence154StirlingNumbers99,100,101,108,109StrongPrime154TriangularNumbers96,134UnitFraction11ValidDigitsSum155-157,157-161,161-163VedamNumbers29Wallis,John173WeakPrime154218FlorentinSmarandacheisanincrediblesourceofideas,onlysomeofwhicharemathematicalinnature.
AmarnathMurthyhaspublishedalargenumberofpapersinthebroadareaof"SmarandacheNotions",whicharemathproblemswhoseorigincanbetracedtoSmarandache.
Thisbookisaneditedversionofmanyofthosepapers,mostofwhichappearedin"SmarandacheNotionsJournal",andmoreinformationaboutSNJisavailableathttp://www.
gallup.
unm.
edu/~smarandache/.
Thetopicscoveredareverybroad,althoughtherearetwomainthemesunderwhichmostofthematerialcanbeclassified.
ASmarandachePartitionFunctionisanoperationwhereasetornumberissplitintopiecesandtogethertheymakeuptheoriginalobject.
Forexample,aSmarandacheRepeatableReciprocalpartitionofunityisasetofnaturalnumberswherethesumofthereciprocalsisone.
Thefirstchapterofthebookdealswithvarioustypesofpartitionsandtheirpropertiesandpartitionsalsoappearinsomeofthelatersections.
Thesecondmainthemeisasetofsequencesdefinedusingvariousproperties.
Forexample,theSmarandachen2nsequenceisformedbyconcatenatinganaturalnumberandits'doubleinthatorder.
Onceasequenceisdefined,thensomepropertiesofthesequenceareexamined.
Acommonexplorationistoaskhowmanyprimesareinthesequenceoraslightmodificationofthesequence.
Thefinalchapterisacollectionofproblemsthatdidnotseemtobeaprecisefitineitheroftheprevioustwocategories.
Forexample,foranynumberd,isitpossibletofindaperfectsquarethathasdigitsumdWhilemanyresultsareproven,alargenumberofproblemsareleftopen,leavingagreatdealofroomforfurtherexploration.
7819319233347ISBN1-931233-34-990000>