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ZHANGWENPENG,LIJUNZHUANG,LIUDUANSEN(EDITORS)RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORY(VOL.
II)PROCEEDINGSOFTHEFIRSTNORTHWESTCONFERENCEONNUMBERTHEORYHexisPhoenix,AZ2005EDITORS:ZHANGWENPENGDEPARTMENTOFMATHEMATICSNORTHWESTUNIVERSITYXI'AN,SHAANXI,P.
R.
CHINALIJUNZHUANG,LIUDUANSENINSTITUTEOFMATHEMATICSSHANGLUOTEACHER'SCOLLEGESHANGLUO,SHAANXI,P.
R.
CHINARESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORY(VOL.
II)PROCEEDINGSOFTHEFIRSTNORTHWESTCONFERENCEONNUMBERTHEORYHexisPhoenix,AZ2005Thisbookcanbeorderedinapaperboundreprintfrom:BooksonDemandProQuestInformation&Learning(UniversityofMicrofilmInternational)300N.
ZeebRoadP.
O.
Box1346,AnnArborMI48106-1346,USATel.
:1-800-521-0600(CustomerService)http://wwwlib.
umi.
com/bod/search/basicPeerReviewers:ShigeruKanemitsu,GraduateSchoolofAdvancedTechnology,UniversityofKinki,11-6Kayanomori,Iizuka820-8555,Japan.
ZhaiWenguang,DepartmentofMathematics,ShandongTeachers'University,Jinan,Shandong,P.
R.
China.
GuoJinbao,CollegeofMathematicsandComputerScience,YananUniversity,Yanan,Shaanxi,P.
R.
China.
Copyright2005byHexisandZhangWenpeng,LiJunzhuang,LiuDuansen,andallauthorsfortheirownarticlesManybookscanbedownloadedfromthefollowingE-LibraryofScience:http://www.
gallup.
unm.
edu/~smarandache/eBooks-otherformats.
htmISBN:1-931233-99-3StandardAddressNumber:297-5092PrintedintheUnitedStatesofAmericaContentsDedicationvPrefacexiOntheSmarandachem-thpowerresidues1ZhangWenpengAnequationinvolvingEuler'sfunction5YiYuanOntheFunction(n)andδk(n)9XuZhefengOnthemeanvalueoftheDirichlet'sdivisorfunctioninsomespecialsets13XueXifengAnumbertheoreticfunctionanditsmeanvalue19RenGanglianOntheadditivek-thpowercomplements23DingLipingOntheSmarandache-Riemannzetasequence29LiJieOnthepropertiesofthehexagon-numbers33LiuYanniOnthemeanvalueofthek-thpowerpartresiduefunction37MaJinpingOntheintegerpartofthem-throotandthek-thpowerfreenumber41LiZhanhu1,2AFormulaForSmarandacheLCMRatioSequence45WangTingOnthem-thpowercomplementssequence47YangHai,FuRuiqinOntheSmarandacheceilfunctionandtheDirichletdivisorfunction51RenDongmeiviiiRESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIOnadualfunctionoftheSmarandacheceilfunction55LuYamingOnthelargestm-thpowernotexceedingn59LiuDuansenandLiJunzhuangOntheSmarandachebackconcatenatedoddsequences63LiJunzhuangandWangNianliangOntheadditivehexagonnumberscomplements71LichaoandYangCundianOnthemeanvalueofanewarithmeticalfunction75YangCundianandLiuDuansenOnthemeanvalueofanewarithmeticalfunction79ZhaoJianandLiChaoOnthesecondclasspseudo-multiplesof5sequences83GaoNanAclassofDirichletseriesanditsidentities87LouYuanbingAnarithmeticalfunctionandtheperfectk-thpowernumbers91YangQianliandYangMingshunAnarithmeticalfunctionanditsmeanvalueformula95RenZhibinandZhaoXiaopengOnthesquarecomplementsfunctionofn!
99FuRuiqin,YangHaiOntheSmarandachefunction103WangYongxingMeanvalueofthek-powercomplementsequences107FengZhiyuOnthem-thpowerfreenumbersequences111ChenGuohuiAnumbertheoreticfunctionanditsmeanvalue115DuJianliOntheadditivek-thpowerpartresiduefunction119ZhangShengshengAnarithmeticalfunctionandthek-thpowercomplements123HuangWeiOnthetrianglenumberpartresidueofapositiveinteger127JiYongqiangContentsixAnarithmeticalfunctionandthek-fullnumbersequences131ZhaoJiantangOnthehybridmeanvalueofsomespecialsequences135LiYanshengandGaoLiOnthemeanvalueofanewarithmeticalfunction143MaJunqingThisbookisdedicatedtoProfessorFlorentinSmarandache,wholistedmanynewandunsolvedproblemsinnumbertheory.
PrefaceArithmeticiswherenumbersrunacrossyourmindlookingfortheanswer.
Arithmeticislikenumbersspinninginyourheadfasterandfasteruntilyoublowupwiththeanswer.
KABOOM!
!
Thenyousitbackdownandbeginthenextproblem.
AlexanderNathansonMathematicsisoftenreferredtoasaservantofscienceandnumbertheoryasaqueenofmathematics,bywhichwearetounderstandthatnumbertheoryshouldcontributetothedevelopmentofsciencebycontinuouslysupplyingchallengingproblemsandillustrativeexamplesofitsowndisciplineinanticipationoflaterpracticaluseintherealworld.
TheresearchonSmarandacheProblemsplaysakeyroleinthedevelopmentofnumbertheory.
Therefore,manymathematiciansshowtheirinterestintheSmarandacheproblemsandtheyconductmuchresearchonthem.
Undersuchcircumstances,wepublishedthebook,Vol.
I,inSeptember,2004.
ThatbookstimulatedmoreChinesemathematicianstopayattentiontoSmarandacheconjectures,openandsolvedproblemsinnumbertheory.
TheFirstNorthwestNumberTheoryConferencewasheldinShangluoTeacher'sCollege,China,inMarch2005.
OneofthesessionswasdedicatedtotheSmarandacheproblems.
Inthatsession,severalprofessorsgaveatalkonSmarandacheproblemsandmanyparticipantslecturedonSmarandacheproblemsbothextensivelyandintensively.
Thisbookincludes34papers,mostofwhichwerewrittenbyparticipantsoftheabovementionedconference.
Allthesepapersareoriginalandhavebeenrefereed.
ThethemesofthesepapersrangefromthemeanvalueorhybridmeanvalueofSmarandachetypefunctions,themeanvalueofsomefamousnumbertheoreticfunctionsactingontheSmarandachesequences,totheconvergencepropertyofsomeinfiniteseriesinvolvingtheSmarandachetypesequences.
Wesincerelythankalltheauthorsandtherefereesfortheirimportantcontributions.
ThanksarealsoduetoDr.
XuZhefengforhiseffortofmakingfilesofLaTeXstyle.
Thelast,butnottheleast,thanksareduetotheteachersandstudentsofShangluoTeacher'sCollegeinChinafortheirgreathelpinoursuccessfulconference.
August10,2005ZhangWenpeng,LiJunzhuang,LiuDuansenONTHESMARANDACHEM-THPOWERRESIDUESZhangWenpengDepartmentofMathematics,NorthwestUniversity,Xi'an,Shaanxi,P.
R.
Chinawpzhang@nwu.
edu.
cnAbstractForanypositiveintegern,letφ(n)betheEulerfunction,andam(n)denotestheSmarandachem-thpowerresiduesfunctionofn.
Themainpurposeofthispaperisusingtheelementarymethodtostudythenumberofthesolutionsoftheequationφ(n)=am(n),andgiveallsolutionsforthisequation.
Keywords:Arithmeticalfunction;Equation;Solutions.
§1.
IntroductionForanypositiveintegern,letn=pα11pα22···pαkkdenotesthefactoriza-tionofnintoprimepowers.
TheSmarandachem-thpowerresiduesfunctionam(n)aredenedasam(n)=pβ11pβ22···pβkk,βi=min{m1,αi},i=1,2.
k.
Inproblem65of[1],ProfessorF.
Smarandacheaskedustostudythepropertiesofthisfunction.
Letφ(n)denotestheEulerfunction.
Thatis,φ(n)denotesthenumberofallpositiveintegersnotexceedingnwhicharerelativelyprimeton.
Itisclearthatφ(n)andam(n)botharemultiplicativefunctions.
Inthispaper,weshallusetheelementarymethodtostudythesolutionsoftheequationinvolvingthesetwofunctions,andgiveallsolutionsforit.
Thatis,weshallprovethefollowing:Theorem.
Letmbeaxedintegerwithm≥2.
Thentheequationφ(n)=am(n)havem+1solutions,namelyn=1,2m,2α3m,α=1,2,m1.
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofthetheorem.
Letn=pα11pα22···pαkkdenotesthefactorizationofnintoprimepowers,thenfromthedenitionsofφ(n)andam(n)wehaveam(n)=pβ11pβ22···pβkk,βi=min{m1,αi},i=1,2.
k(1)ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(10271093,60472068)2RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIandφ(n)=pα111(p11)pα212(p21)···pαk1k(pk1).
(2)Itisclearthatn=1isasolutionoftheequationφ(n)=am(n).
Ifn>1,thenwewilldiscusstheprobleminfourcases:(i)Ifαi>m,thenwehaveαi1≥m,max1≤j≤k{βj}≤m1.
Combining(1)and(2),weknowthatφ(n)=am(n).
Thatis,thereisnoanysolutionsatisedφ(n)=am(n)inthiscase;(ii)Ifα1=α2αk=m,thenwehaveam(n)=pm11pm12···pm1kandφ(n)=pm11(p11)pm12(p21)···pm1k(pk1).
Itisclearthatonlyn=2misasolutionoftheequationinthiscase;(iii)Ifmax1≤i≤k{αi}αk1,weknowthatφ(n)=am(n)forallninthiscase;(iv)Ifi,jsuchthatαi=mandαj1,from(3)wecandeducethatonlyonetermwiththeformpi1intheleftsideof(3).
Thatis,n=2αpm.
Fromφ(2αpm)=am(2αpm)pm1(p1)=2pm1,wegetn=2α3m,1≤α12,wehavetheidentity∞n=1n∈A1ns=ζ(2s)ζ(3s)ζ(6s),whereζ(s)istheRiemannzeta-function.
Takings=1and2,andnotethatζ(2)=π2/6,ζ(4)=π4/90,ζ(6)=π6/945,ζ(12)=691π12/638512875,wemayimmidiatelydeducethefol-lowingidentities:∞n=1n∈A1n=3152π4ζ(3)and∞n=1n∈A1n2=1501513821π2.
ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(10271093)6RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII§2.
ProofofthetheoremInthissection,wewillcompletetheproofofthetheorem.
Firstnotethatbothφ(n)andJ(n)aremultiplicativefunctions,andifn=pα,thenJ(p)=p2,φ(p)=p1,J(pα)=pα2(p1)2andφ(pα)=pα1(p1)forallpositiveintegerα>1andprimep.
Sowewilldiscussthesolutionsoftheequationφ2(n)=nJ(n)intothreecases.
(a)Itisclearthatn=1isasolutionoftheequationφ2(n)=nJ(n).
(b)Ifn>1,letn=pα11pα22···pαssdenotestheprimepowersdecomposi-tionofnwithallαi≥2(i=1,2,k),thenwehaveJ(n)=pα121(p11)2···pαk2k(pk1)2andφ2(n)=pα111(p11)pα212(p21)···pαk1k(pk1)2.
Forthiscase,nisalsoasolutionoftheequationφ2(n)=nJ(n).
(c)Ifn=p1p2···prpαr+1r+1···pαkkwithp11,j=r+1,r+2,k,thenfrom(b)andthedenitionofφ(n)andJ(n)wehaveφ2(n)=nJ(n)ifandonlyifφ2(p1p2···pr)=p1p2···prJ(p1p2···pr)or(p11)2(p21)2···(pr1)2=p1p2···pr(p12)(p22)pr2).
Itisclearthatprcannotdivide(p11)2(p21)2···(pr1)2.
Sotheequationφ2(n)=nJ(n)hasnosolutioninthiscase.
Combiningtheabovethreecaseswemayimmediatelyobtainthesetofallsolutionsoftheequationφ2(n)=nJ(n)isn=pα11pα22···pαsswithallαi>1and1.
Thatis,Aisthesetofallsquare-fullnumbersand1.
Nowwedenethearithmeticalfunctiona(n)asfollows:a(n)=1,ifn∈A,0,ifotherwise.
Foranyrealnumbers>0,itisclearthat∞n=1n∈A1ns1.
Sofors≥2,fromtheEulerproductformula(see[2]),wehaveAnequationinvolvingEuler'sfunction17∞n=11ns=p1+a(p2)p2s+a(p3)p3s+···=p1+1p2s+1p3s+···=p1+1p2s(11ps)=pp2sps+1p2sps=pp3s+1ps(p2s1)=pp6s1ps(p2s1)(p3s1)=p11p6s(11p2s)(11p3s)=ζ(2s)ζ(3s)ζ(6s),whereζ(s)istheRiemannzeta-function,andpdenotestheproductoverallprimes.
Thiscompletestheproofofthetheorem.
References[1]F.
Smaradache.
Onlyproblems,notsolutions,XiquanPublishingHouse,Chicago,1993.
[2]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
ONTHEFUNCTION(N)ANDδK(N)XuZhefengDepartmentofMathematics,NorthwestUniversity,Xi'an,Shaanxi,P.
R.
Chinazfxu@nwu.
edu.
cnAbstractThemainpurposeofthispaperisusingtheelementarymethodtostudythedivisibilityofδk(n)by(n),andndallthensuchthat(n)|δk(n).
Keywords:Eulerfunction;Arithmeticalfunction;Divisibility.
§1.
IntroductionandmainresultsForaxedpositiveintegerkandanypositiveintegern,wedeneanewarithmeticfunctionasfollowing:δk(n)=max{d|d|n,(d,k)=1}Ifn≥1theEulerfunction(n)isdenedtobethenumberofpositiveintegersnotexceedingnwhicharerelativelyprimeton;thus,(n)=nk=11wherenk=1indicatesthatthesumisextendedoverthosekrelativelyprimeton.
Inthispaper,weshallstudythedivisibilityofδk(n)by(n),andndallthensuchthat(n)|δk(n).
Infact,weshallprovethefollowingresult:Theorem.
(n)|δk(n)ifandonlyifn=2α3β,whereα>0,β≥0,α,β∈N.
§2.
ProofofthetheoremInthissection,wewillcompletetheproofofthetheorem.
FirstweneedthefollowingLemma.
ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(60472068)10RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIILemmaForn≥1,wehave(n)=np|n11p.
Proof.
SeeTheorem2.
4inreference[2].
NowweusetheabovelemmatocompletetheproofofTheorem.
Wewilldiscussitintwocases.
I.
(n,k)=1,thenδk(n)=n.
Letn=pα11pα22···pαssisthefactorizationofnintoprimepowers.
FromtheLemma,wecanwrite(n)=pα111(p11)pα212(p21)···pαs1s(ps1),If(n)|δk(n),thenpα111(p11)pα212(p21)···pαs1s(ps1)|n,thatis(p11)(p21)ps1)|p1p2···ps,fromtheaboveformula,wearesureofi)p1=2,ifnot,(p11)(p21)ps1)isevenbutp1p2···psoddandii)s≤2,since2|(pi1)(i=2,3,s)butp2p3···pscannotbedividedby2.
Ifs=1,thenn=2α,α≥1.
Ifs=2,thenn=2α3β,α≥1,β>1.
Thus,wecanobtainthatn=2α3β,α≥1,β≥0suchthat(n)|δk(n)when(n,k)=1.
II.
(n,k)=1,wecanwriten=n1·n2,where(n1,k)=1and(n1,n2)=1,thenδk(n)=n1,and(n)=(n1)(n2),If(n)|δk(n),thatmeans(n1)(n2)|n1,thatis(n1)|n1(1)(n2)|n1(2)from(1),wecangetn1=2α13β1,(3)OntheFunction(n)andδk(n)111whereα1≥1,β1≥0,α1,β1∈N.
combining(3)and(4),wecaneasilyget(n2)|2α13β1,thismeans(n2)=1.
Otherwise(n1,n2)=1.
Son2=1Altogether,whether(n,k)=1ornot,wecanobtainn=2α3β,α≥1,β≥0suchthat(n)|δk(n).
Thiscompletestheproofofthetheorem.
References[1]Ibstedt.
SurnigOntheOceanofNumbers-AFewSmarandacheNo-tionsandSimilarTopics,ErhusUniversityPress,NewMexico.
[2]T.
M.
Apostol.
IntroductiontoAnalyticNumberTheory,Springer-verlag,NewYork,1976.
[3]F.
Mertens.
EinBeitragZurAnalytischenZahlentheorie,Crelle'sJour-nal,78(1874),46-62.
ONTHEMEANVALUEOFTHEDIRICHLET'SDIVISORFUNCTIONINSOMESPECIALSETSXueXifengDepartmentofMathematics,NorthwestUniversity,Xi'an,Shaanxi,P.
R.
ChinaAbstractThemainpurposeofthispaperisusingtheanalyticmethodstostudythemeanvaluepropertiesoftheDirichlet'sdivisorfunctioninsomespecialsets,andgiveseveralinterestingasymptoticformulaeforthem.
Keywords:Divisorfunction;Meanvalue;Asymptoticformula.
§1.
IntroductionForanypositiveintegernandk≥2,thek-thpowercomplementfunctionbk(n)ofnisthesmallestpositiveintegersuchthatnbk(n)isaperfectk-thpower.
Inproblem29ofreference[1],ProfessorF.
Smarandacheaskedustostudythepropertiesofthisfunction.
Aboutthisproblem,somepeoplehadstudieditbefore,andobtainedsomeinterestingconclusions,seereferences[4]and[5].
Inthispaper,wedenetwonewsetsB={n∈N,bk(n)|n}andC={n∈N,n|bk(n)}.
ThenweusetheanalyticmethodstostudythemeanvaluepropertiesoftheDirichlet'sdivisorfunctiond(n)actingonthesetwospecialsets,andobtaintwointerestingasymptoticformulaeforthem.
Thatis,weshallprovethefollowing:Theorem1.
Foranyrealnumberx≥1,wehavetheasymptoticformulan≤xn∈Bd(n)=mx1mζm+1(2)Rp1m·f(logx)++Ox12m+ε,whereRp1m=p1+pm(p1m1)(m+1)+p1m(p+1)m+1(p1m1)2p1m12m+1i=2m+1ipm+1i(p+1)m+1(p1m1)2,ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(60472068).
14RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIf(y)isapolynomialofywithdegreem=[k+12],andεisanyxedpositivenumber.
Theorem2.
Foranyrealnumberx≥1,wehavetheasymptoticformulan≤xn∈Cd(n)=xlogxζ(l+1)p1(l+1)(p1)pl+2p+Ax++O(x12+ε),wherel=k2,Aisaconstant,andεisanyxedpositivenumber.
§2.
ProofofthetheoremsInthissection,weshallcompletetheproofofthetheorems.
Infact,letn=pα11pα22···pαssdenotesthefactorizationofnintoprimepowers.
Thenitisclearthatbk(n)=bk(pα11pα22···pαss)=bk(pα11)bk(pα22)···bk(pαss).
Thatis,bk(n)isamultiplicativefunction.
Sowerststudytheprobleminthecasen=pα.
(1)Ifα≥k,thenfromthedenitionofbk(n)weknowthatbk(n)|n.
Thereforen∈B.
(2)Ifα≤k,thenbk(n)=pkα.
Socombining(1),(2)andthedenitionofBandC,wededucethatn∈Bifα≥[k+12],andn∈Cifα≤[k2].
Nowweprovethetheorem1andtheorem2respectively.
Firstletf(s)=n∈Bd(n)ns.
Thenfromthedenitionofbk(n)andB,thepropertiesoftheDirichlet'sdivisorfunctionandtheEulerproductformula[2],wehavef(s)=p1+d(pm)pms+d(pm+1)p(m+1)s+···=p1+m+1pms+m+2p(m+1)s+···=p1+m+1pms+m+1pms(ps1)+pspms(ps1)2=ζm+1(ms)ζm+1(2ms)p1+pm2s((ps1)(m+1)+ps)(pms+1)m+1(ps1)2(1)(ps1)2m+1i=2m+1ipm(m+1i)s(pms+1)m+1(ps1)2,OnthemeanvalueoftheDirichlet'sdivisorfunctioninsomespecialsets115whereζ(s)istheRiemannzeta-functionandm=[k+12].
Obviously,wehave|d(n)|≤n,∞n=1d(n)nσ≤1σ11m,whereσ>11mistherealpartofs.
ThereforebyPerronformula[3],withs0=0,b=2m,T=x32m,wehaven≤xn∈Bd(n)=12πi2m+iT2miTζm+1(ms)ζm+1(2ms)R(s)xssds+O(x12m+ε),whereR(s)=p1+pm2s((ps1)(m+1)+ps)(ps1)2m+1i=2m+1ipm(m+1i)s(pms+1)m+1(ps1)2.
Toestimatethemainterm12πi2m+iT2miTζm+1(ms)ζm+1(2ms)R(s)xssds,wemovetheintegrallinefroms=2m±iTtos=12m±iT,thenthefunctionζm+1(ms)ζm(2ms)R(s)xsshaveonem+1orderpolepointats=1mwithresiduelims→11m!
(ms1)m+1ζm+1(ms)R(s)xsζm+1(2ms)s(m)=lims→11m!
(m0)(ms1)m+1ζm+1(ms)(m)R(s)xsζm+1(2ms)s++lims→11m!
(m1)(ms1)m+1ζm+1(ms)(m1)R(s)xsζm+1(2ms)s+···+lims→11m!
(mm)(ms1)m+1ζm+1(ms)R(s)xsζm+1(2ms)s(m)=mx1mζm+1(2)R1mf(logx)+Ox12m+,wheref(y)isapolynomialofywithdegreek,andεisanyxedpositivenumber.
16RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIISothatwecanget12πi2m+iT2miT+12m+iT2m+iT+12miT12m+iT+2miT12miTζm+1(ms)xsζm+1(2ms)sR(s)ds=mx1mζm+1(2)R1mf(logx)+O(x12m+).
Itiseasytoestimate12πi12m+iT2m+iTζm+1(ms)xsζm+1(2ms)sR(s)dsx12m+ε,12πi2miT12miTζm+1(ms)xsζm+1(2ms)sR(s)dsx12m+ε,and12πi12miT12m+iTζm+1(ms)xsζm+1(2ms)sR(s)dsx12m+ε.
Therefore,wehaven≤xn∈Bd(n)=mx1mζm+1(2)Rp1m·f(logx)++Ox12m+ε,whereRp1m=p1+pm(p1m1)(m+1)+p1mp1m12m+1i=2m+1ipm+1i(p+1)m+1(p1m1)2.
ThiscompletestheproofofTheorem1.
Foranyintegerk≥2andanyrealnumbers>1,letg(s)=n∈Cd(n)ns.
ThenfromtheEulerproductformula[3]wehaveg(s)=n∈Cd(n)ns=p1+d(p)ps+d(p2)p2sd(pl)pmsOnthemeanvalueoftheDirichlet'sdivisorfunctioninsomespecialsets217=p1+2ps+3p2sl+1pls=p111ps1+1ps+1p2s+1p3s1plsl+1p(l+1)s=ζ(s)p11p(l+1)s11psl+1p(l+1)s=ζ2(s)ζ((l+1)s)p1(l+1)(ps1)p(l+2)sps,wherel=[k2].
SobyPerronformula[3]andthemethodsofprovingTheorem1wecaneasilygetTheorem2.
References[1]F.
Smarandache.
OnlyProblems,Notsolutions.
Chicago:XiquanPub-lishingHouse,1993.
[2]ApostolTM.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
[3]PanChengdongandPanChengbiao.
FoundationofAnalyticNumberTheory.
Beijing:Sciencepress,1997.
[4]ZhuWeiyi.
Onthek-powercomplementandk-powerfreenumberse-quence.
SmarandacheNotionsJournal,2004,14:66-69.
[5]LiuHongyanandLouYuanbing.
ANoteonthe29-thSmarandache'sproblem.
SmarandacheNotionsJournal,2004,14:156-158.
ANUMBERTHEORETICFUNCTIONANDITSMEANVALUERenGanglianDepartmentofMathematics,NorthwestUniversity,Xi'an,Shaanxi,P.
R.
ChinaAbstractLetpandqbetwodistinctprimes,epq(n)denotethelargestexponentofpowerpqwhichdividesn.
Inthispaper,westudythepropertiesofthesequenceepq(n),andgiveaninterestingasymptoticformulaforthemeanvaluen≤xepq(n).
Keywords:Largestexponent;Asymptoticformula;Meanvalue.
§1.
IntroductionLetpandqbetwodistinctprimes,epq(n)denotethelargestexponentofpowerpqwhichdividesn.
Inproblem68of[1],ProfessorF.
Smarandacheaskedustostudythepropertiesofthesequenceep(n).
Aboutthisproblem,somepeoplehadstudiedit,andobtainedaseriesofinterestingresults(Seereferences[4]).
Inthispaper,weusetheanalyticmethodtostudytheproper-tiesofthesequenceepq(n),andgiveasharpasymptoticformulaforitsmeanvaluen≤xepq(n).
Thatis,weshallprovethefollowing:Theorem.
Letpandqbetwodistinctprimes,thenforanyrealnumberx≥1,wehavetheasymptoticformulan≤xepq(n)=xpq1+Ox1/2+ε,whereεisanyxedpositivenumber.
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofTheorem.
Foranycomplexs,wedenethefunctionf(s)=∞n=1epq(n)ns.
ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(60472068).
20RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIINotethatthedenitionofepq(n),andapplyingtheEulerproductformula(SeeTheorem11.
6of[2]),wemaygetf(s)=∞α=0∞t=1∞n1=1(n1,pq)=1α(pα+tqαn1)s+∞α=0∞t=1∞n1=1(n1,pq)=1α(pαqα+tn1)s+∞α=0∞n1=1(n1,pq)=1α(pαqαn1)s=∞α=0α(pq)αs∞t=11pts∞n1=1(n1,pq)=11n1s+∞α=0α(pq)αs∞t=11qts∞n1=1(n1,pq)=11n1s+∞α=0α(pq)αs∞n1=1(n1,pq)=11n1s=ζ(s)(pq)s1,whereζ(s)istheRiemannzeta-function.
Obviously,wehaveepq(n)≤logpqn≤lnn∞n=1epq(n)nσ≤1σ1,whereσistherealpartofs.
ThereforebyParron'sformula(Seereference[3])wecangetn≤xepq(n)ns0=12πib+iTbiTζ(s+s0)(pq)s+s01xssds+OxbB(b+σ0)T+Ox1σ0H(2x)min1,logxT+Oxσ0H(N)min1,xxwhereNisthenearestintegertox,x=|xN|.
Takings0=0,b=32,H(x)=lnx,B(σ)=1σ1,wehaven≤xepq(n)=12πi32+iT32iTζ(s)(pq)s1xssds+O(x32+εT).
Toestimatethemainterm12πi32+iT32iTζ(s)(pq)s1xssds,Anumbertheoreticfunctionanditsmeanvalue121wemovetheintegrallinefroms=32±iTtos=12±iT.
Thistime,thefunctiong(s)=ζ(s)(pq)s1xsshasasimplepolepointats=1,andtheresidueisxpq1.
Sowehave12πi32+iT32iT+12+iT32+iT+12iT12+iT+32iT12iTζ(s)(pq)s1xssds=xpq1.
TakingT=x,andnotethat12πi12+iT32+iT+32iT12iTζ(s)(pq)s1xssds3212ζ(σ+iT)1(pq)σ1x32Tdσx32+εT=x12+εand12πi12iT12+iTζ(s)(pq)s1xssdsT0|ζ(12+it)1(pq)121x12t|dtx12+ε,wemayimmediatelygettheasymptoticformulan≤xepq(n)=xpq1+Ox1/2+ε.
ThiscompletestheproofofTheorem.
References[1]F.
Smarandache.
OnlyProblems,Notsolutions.
Chicago:XiquanPub-lishingHouse,1993.
[2]ApostolTM.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
[3]PanChengdongandPanChengbiao.
FoundationofAnalyticNumberTheory.
Beijing:Sciencepress,1997.
[4]ZhangWenpeng.
ReseachonSmarandacheproblemsinnumbertheory.
HexisPublishingHouse,2004.
ONTHEADDITIVEK-THPOWERCOMPLEMENTSDingLipingDepartmentofMathematics,NorthwestUniversity,Xi'an,Shaanxi,P.
R.
ChinaAbstractThemainpurposeofthispaperisusingtheelementaryandanalyticmethodstostudythemeanvaluepropertiesoftheadditivek-thpowercomplements,andgivesomeinterestingasymptoticformulaeforit.
Keywords:Additivek-thpowercomplements;Meanvalue;Asymptoticformula.
§1.
IntroductionForanypositiveintegern,theSmarandachek-thpowercomplementsbk(n)isthesmallestpositiveintegersuchthatnbk(n)isacompletek-thpower,seeproblem29of[1].
SimilartotheSmarandachek-thpowercomplements,theadditivek-thpowercomplementsak(n)isdenedasfollows:ak(n)isthesmallestnonnegativeintegersuchthatak(n)+nisaperfectk-thpower.
Forexample,ifk=2,wehavetheadditivesquarecomplementssequence{a2(n)}(n=1,2,asfollows:a2(1)=0,a2(2)=2,a2(3)=1,a2(4)=0,a2(5)=4,a2(6)=3,a2(7)=2,a2(8)=1,a2(9)=0,Aboutthisproblem,manyauthorshavestudieditbefore,andobtainedsomeinterestingresults.
Forexample,Z.
F.
Xu[4]studiedthemeanvaluepropertiesoftheadditivek-thpowercomplements,andgavethefollowing:Proposition.
Foranyrealnumberx≥3andxedpositiveintegerk≥2,wehavetheasymptoticformula:n≤xak(n)=k24k2x21k+Ox22k.
Foranyxedpositiveintegerm,thedenitionofthearithmeticalfunctionδm(n)isδm(n)=max{d∈N|d|n,(d,m)=1},ifn=0,0,ifn=0.
Inthispaper,weshallusetheelementaryandanalyticmethodstostudythemeanvaluepropertiesofthenewarithmeticalfunctionδm(ak(n)),andgiveThisworkissupportedbyN.
S.
F.
ofP.
R.
China(60472068).
24RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIaninterestingasymptoticformulaforit.
Thatis,weshallprovethefollowingconclusion:Theorem.
Foranyrealnumberx≥3andpositiveintegerm,wehavetheasymptoticformula:n≤xδm(ak(n))=k22(2k1)x21kp|mpp+1+Ox22k,ifk>3,wherep|mdenotestheproductoverallprimedivisorspofm,andisanyxedpositivenumber.
Takingk=2,3inourTheorem,wemayimmediatelydeducethefollowing:Corollary.
Foranyrealnumberx≥1,wehavetheasymptoticformulaen≤xδm(a2(n))=23x32p|mpp+1+Ox54+andn≤xδm(a3(n))=910x53p|mpp+1+Ox43+.
§2.
SomelemmasTocompletetheproofofthetheorem,weneedfollowingLemmas:Lemma1.
Foranyrealnumberx>1andpositiveintegerm,wehavetheasymptoticformulan≤xδm(n)=x22p|mpp+1+Ox32+,whereisanyxedpositivenumber.
Proof.
Lets=σ+itbeacomplexnumberandf(s)=∞n=1δm(n)ns.
Notethatδm(n)n,soitisclearthatf(s)isaDirichletseriesabsolutelyconvergentforRe(s)>2,bytheEulerproductformula[2]andthedenitionofδm(n)wegetf(s)=∞n=1δm(n)ns=p1+δm(p)ps+δm(p2)p2sδm(p2n)pns+···=p|m1+δm(p)ps+δm(p2)p2sδm(p2n)pns+···Ontheadditivek-thpowercomplements125*pm1+δm(p)ps+δm(p2)p2sδm(p2n)pns+···=p|m1+1ps+1p2s1pns+···*pm1+pps+p2p2sp2npns+···=p|m111pspm111ps1=ζ(s1)p|mpspps1,(1)whereζ(s)istheRiemannzeta-function,andpdenotestheproductoverallprimes.
From(1)andPerron'sformula[3],wehaven≤xδm(n)=12πi52+iT52iTζ(s1)p|mpspps1·xssds+Ox52+T,(2)whereisanyxedpositivenumber.
Nowwemovetheintegrallinein(2)froms=52±iTtos=32±iT.
Thistime,thefunctionζ(s1)p|mpspps1·xsshasasimplepolepointats=2withresiduex22p|mpp+1.
(3)Hence,wehave12πi52iT32iT+52+iT52iT+32+iT52+iT+32iT32+iTζ(s1)p|mpspps1·xssds=x22p|mpp+1.
(4)Wecaneasilygettheestimate12πi52iT32iT+32+iT52+iTζ(s1)p|mpspps1·xssdsx52+T(5)and12πi32iT32+iTζ(s1)p|mpspps1·xssdsx32+.
(6)26RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIITakingT=x,combining(2),(4),(5)and(6)wededucethatn≤xδm(n)=x22p|mpp+1+O(x32+).
(7)ThiscompletestheproofofLemma1.
Lemma2.
Foranyrealnumberx≥3andanynonnegativearithmeticalfunctionf(n)withf(0)=0,wehavetheasymptoticformula:n≤xf(ak(n))=x1k1t=1n≤g(t)f(n)+On≤gx1kf(n),where[x]denotesthegreatestintegernotexceedingxandg(t)=k1i=1ikti.
Proof.
Seereference[4].
§3.
ProofofthetheoremInthissection,wewillcompletetheproofofthetheorem.
Fromthedeni-tionofδm(ak(n)),Lemma1andLemma2,wehaven≤xδm(ak(n))=x1k1t=1n≤g(t)δm(n)+On≤gx1kδm(n)=x1k1t=1k2t2k22p|mpp+1+O(t2k3)+Ox11klnxifk>3=k22p|mpp+1x1k1t=1t2k2+O(x22k)=k22(2k1)p|mpp+1x21k+O(x22k).
Forthecasesofk=2,3,wecanalsoprovetheresultsbyLemma2.
Forexample,n≤xδm(a2(n))Ontheadditivek-thpowercomplements227=x121t=1n≤2tδm(n)+On≤gx12δm(n)=x121t=12t2p|mpp+1+O(t32+)+Ox112lnx=2x323p|mpp+1+O(x54+).
Thiscompletestheproofofthetheorem.
References[1]F.
Smarandache.
Onlyproblems,notsolutions,XiquanPublishingHouse,Chicago,1993.
[2]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
[3]PanChengdongandPanChengbiao.
FoundationofAnalyticNumberTheory.
Beijing:SciencePress,1997.
[4]XuZhefeng.
Ontheadditivek-thpowercomplements.
ResearchonSmarandacheProblemsinNumberTheory,Hexis13-16,2004.
ONTHESMARANDACHE-RIEMANNZETASEQUENCELiJieDepartmentofMathematics,NorthwestUniversity,Xi'an,Shaanxi,P.
R.
Chinajli0815@126.
comAbstractThemainpurposeofthispaperisusingtheelementarymethodtostudythepropertiesoftheSmarandache-Riemannzetasequence,andsolveaconjectureposedbyMurthy.
Keywords:Smarandache-Riemannzetasequence;Murthy'sconjecture;Elementarymethod.
§1.
IntroductionForanypositiveintegern,weleta(2n,2)denotesthesumofthebase2digitsof2n.
Thatis,if2n=a12α1+a22α2as2αswithαs>αs1>.
.
.
>α1≥0,whereai=0or1,i=1,2,s,thena(2n,2)=si=1ai.
ForanycomplexnumberswithRe(s)>1,wedeneRiemann-zetafunctionasζ(s)=∞n=11ns.
Foranypositiveintegern,letTnbeanumbersuchthatζ(2n)=π2nTn,whereπisratioofthecircumferenceofacircletoitsdiameter.
ThenthesequenceT={Tn}∞n=1iscalledtheSmarandache-Riemannzetasequence.
In[1],MurthybelievedthatTnisasequenceofintegers.
Simultaneous,heproposedthefollowing:Conjecture.
NotwotermsofTnarerelativelyprime.
Inthispaper,weshallprovethefollowingconclusion:Theorem.
ThereexistsinnitepositiveintegersnsuchthatTnisnotaninteger.
FromthisTheoremweknowthattheMurthy'sconjectureisnotcorrect,becausethereexistsinnitepositiveintegersnsuchthatTnisnotaninteger.
30RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII§2.
SomesimplelemmasBeforetheproofofthetheorem,somesimplelemmaswillbeuseful.
Lemma1.
Iford(2,(2n)!
)αs.
Sofromthisformulawehaveα2(2n)≡+∞i=12n2i=+∞i=1a12α1+a22α2as2αs2i=sj=1αjk=1aj2αjk=sj=1aj(1+2+222αj1)=sj=1aj·2αj121=sj=1(aj2αjaj)=2na(2n,2)ThiscompletestheproofofLemma2.
§3.
ProofofthetheoremInthissection,weshallcompletetheproofofthetheorem.
FromLemma2andtheTheorem3.
14of[3]weknowthatord(2,(2n)!
)=α2(2n)≡+∞i=12n2i=2na(2n,2).
Nowitisclearthatord(2,(2n)!
)α3>α2>α1≥0,thena(2n,2)≥3.
Forthesen,fromLemma1weknowthatTnisnotaninteger.
Sincethereareinnitepositiveintegersα1,α2,α3andα4,itmeansthatthereexistsinnitepositiveintegersnsuchthata(2n,2)≥3.
Therefore,thereexistsinnitepositiveintegersnsuchthatTnisnotaninteger.
ThiscompletestheproofofTheorem.
References[1]A.
Murthy,Somemoreconjecturesonprimesanddivisors,SmarandacheNotionsJ.
12(2001),pp.
311.
[2]LeMaohua,TheSmarandache-Riemannzetasequence,SmarandacheNotionsJ.
14(2004),pp.
346.
[3]TomM.
Apostol,IntroductiontoAnalyticNumberTheory,NewYork:Springer-Verlag,1976.
ONTHEPROPERTIESOFTHEHEXAGON-NUMBERSLiuYanniDepartmentofMathematics,NorthwestUniversity,Xi'an,Shaanxi,P.
R.
ChinaAbstractInthispaper,weshallusetheelementarymethodtostudythepropertiesofthehexagon-numbers,andgiveaninterestingidentityinvolvingthehexagon-numbers.
Keywords:Hexagon-number;Arithmeticalproperty;Identity.
§1.
IntroductionForanypositiveintegerm,ifn=m(2m1),thenwecallsuchanintegernashexagon-number(seereference[1]).
Thehexagon-numbers1,6,15,28,···arecloselyrelatedtothehexagons.
Moreover,thehexagon-numbersarethepartialsumsofthetermsinthearithmeticprogression1,5,9,13,4n+1,Foranypositiveintegern,letmbethelargestpositiveintegerthatsatisfytheinequalitym(2m1)≤n1,theinnityseriesf(s)isconver-gent,andf(2)=∞n=11a2(n)=53π24ln2.
§2.
ProofofTheoremInthissection,weshallcompletetheproofofTheorem.
Firstfromthedenitionofa(n)weknowthatthereexists(m+1)(2m+1)m(2m1)=4m+134RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIsolutionsfortheequationa(n)=m(2m1).
Sowecaneasydeducethatf(s)=∞n=11as(n)=∞m=1∞n=1a(n)=m(2m1)1as(n)=∞m=14m+1ms(2m1)s.
Thuswemayconcludethatf(s)isconvergentif2s1>1.
Thatis,s>1.
Nowweshallusetheelementarymethodtocalculatetheexactvalueoff(2).
Applyingtheaboveidentitywehavef(2)=∞m=14m+1m2(2m1)2=∞m=11m2+∞m=112(2m1)2∞m=18m(2m1)=∞m=11m2+12∞m=11m2∞m=114m2∞m=18m(2m1)=ζ(2)+12ζ(2)14ζ(2)∞m=18m(2m1)=10ζ(2)∞m=18m(2m1)=10ζ(2)4∞m=112m1∞m=112m=10ζ(2)4112+1314+1516whereζ(s)istheRiemannzeta-function.
FromtheTaylor'sexpansion(seereference[2])forln(1+x)weknowthatln(1+x)=xx22+x33x44xnnUsingthisidentitywithx=1wemayimmediatelygetln2=112+1314+1516Combingalltheabove,andnotethatζ(2)=π26wemayobtainf(2)=10ζ(2)4ln2=53π24ln2.
ThiscompletestheproofofTheorem.
Onthepropertiesofthehexagon-numbers35References[1]TomM.
Apostol,IntroductiontoAnalyticNumberTheory,NewYork:Springer-Verlag,1976.
[2]TomM.
Apostol,MathematicalAnalysis(2ndEdition),Addison-WesleyPublishingCo.
,Reading,Mass.
-London-DonMills,Ont.
,1974.
ONTHEMEANVALUEOFTHEK-THPOWERPARTRESIDUEFUNCTIONMaJinpingDepartmentofMathematics,NorthwestUniversity,Xi'an,Shaanxi,P.
R.
ChinaAbstractSimilartotheSmarandachek-thpowercomplements,wedenethek-thpowerpartresiduefunctionfk(n)isthesmallestnonnegativeintegersuchthatnfk(n)isaperfectk-thpower.
Themainpurposeofthispaperisusingtheelementarymethodstostudythemeanvaluepropertiesof1fk(n)+1,andgiveaninterestingasymptoticformulaforit.
Keywords:k-thpowerpartresiduefunction;Meanvalue;Asymptoticformula.
§1.
IntroductionandresultsForanypositiveintegern,theSmarandachek-thpowercomplementsbk(n)isthesmallestpositiveintegersuchthatnbk(n)isaperfectk-thpower(seeproblem29of[1]).
SimilartotheSmarandachek-thpowercomplements,XuZhefengin[2]denedtheadditivek-thpowercomplementsak(n)asfollows:ak(n)isthesmallestnonnegativeintegersuchthatn+ak(n)isaperfectk-thpower.
Asageneralizationof[2],wewilldenethek-thpowerpartresiduefunc-tionfk(n)asthesmallestnonnegativeintegersuchthatnfk(n)isaperfectk-thpower.
Forexample,ifk=2,wehavethesquarepartresiduesequence{f2(n)}(n=1,2,asfollowing:0,1,2,0,1,2,3,4,0,1,2,3,4,5,6,0,1,2,.
.
.
.
Meanwhile,letpbeaprime,ep(n)denotesthelargestexponentofpowerpwhichdividesn.
Abouttherelationsbetweenep(n)andfk(n),itseemsthatnonehadstudiedthembefore,atleastwecouldn'tndanyreferenceaboutit.
Inthispaper,weusetheelementarymethodstostudythemeanvalueprop-ertiesof1fk(n)+1andep(fk(n)),andobtaintwosharperasymptoticformulaeforthem.
Thatis,wewillprovethefollowingconclusions:Theorem1.
Foranyrealnumberx≥3,wehavetheasymptoticformulan≤x1fk(n)+1=k1kx1klnx+(lnx+γk+1)x1k+O(lnx),whereγistheEulerconstant.
38RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIITheorem2.
Foranyrealnumberx≥3,wehavetheasymptoticformulan≤xep(fk(n))=1p1x+Okp1x11k.
§2.
SomeLemmasTocompletetheproofoftheabovetheorems,weneedfollowingseveralLemmas.
Lemma1.
Foranyrealnumberx≥1,wehaven≤xep(n)=1p1x+O(ln2x).
Proof.
Seereference[3].
Lemma2.
Leth(n)beanonnegativearithmeticalfunctionwithh(0)=0.
Then,foranyrealnumberx≥1wehavetheasymptoticformula:n≤xh(fk(n))=M1t=1n≤g(t)h(n)+On≤g(M)h(n),whereg(t)=k1i=1kitiandM=[x1k],[x]denotesthegreatestintegernotexceedingx.
Proof.
Foranyrealnumberx≥1,letMbeaxedpositiveintegersuchthatMk≤x1,wehavetheasymptoticformulan≤x1n=lnx+γ+O1x,whereγistheEulerconstant.
Proof.
Seereference[4].
Lemma4Foranyrealnumberx≥3,wehavetheasymptoticformulan≤xfk(n)=k22(2k1)x21k+Ox22k.
Proof.
Leth(n)=nandM=[x1k],thenfromLemma2andEulersum-mationformula(seereference[4])weobtainn≤xfk(n)=M1t=1n≤g(t)n+On≤g(M)n=12[x1k]1t=1k2t2k2+O(x22k)=k22(2k1)x21k+O(x22k).
ThisprovesLemma4.
§3.
ProofofthetheoremsInthissection,weshallcompletetheproofofTheorems.
FirstweproveTheorem1.
Leth(n)=nandM=[x1k],thenfromLemma2andLemma3weobtainn≤x1fk(n)+1=M1t=1n≤g(t)1n+1+On≤g(M)1n+1=M1t=1ln(ktk1)+ln1+O1t+γ+O1g(t)+O(lnx)=(k1)ln((M1)!
)+(lnk+γ)(M1)+O(lnx)=(k1)([x1k]1)ln([x1k]1)([x1k]1)+(lnk+γ)([x1k]1)+O(lnx)=k1kx1klnx+(lnx+γk+1)x1k+O(lnx).
40RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIThiscompletestheproofofTheorem1.
TheproofofTheorem2.
Notethatthedenitionofep(n),fromLemma1andLemma4wehaven≤xep(fk(n))=Mt=1(t1)k≤n1,thenfromtheaboveformulawehavedk|n(d)=dk|uk(d)=d|u(d)=1u=0=a(n).
Ontheintegerpartofthem-throotandthek-thpowerfreenumber43ThisprovesLemma2.
§3.
ProofofthetheoremInthissection,weshallcompletetheproofofthetheorem.
Foranyrealnumberx≥1andpositiveintegerk≥2,fromLemma1andLemma2andthedenitionofbm(n)wehaven≤xn∈Akbm(n)=n≤xa(n)bm(n)=n≤xdk|n(d)n1m=dkt≤x(d)(dkt)1m=dk≤x(d)dkmt≤x/dkt1m+Odk≤xt≤x/dk1=dk≤x(d)dkmmm+1xm+1m1dk(m+1)m+Oxdk1m+Oxdk≤x1dk=mm+1xm+1mdk≤x(d)dk+Ox1mdk≤x1+O(x)=1ζ(k)mm+1xm+1m+O(x).
ThiscompletestheproofofTheorem.
NowtheCorollariesfollowsfromζ(2)=π2/6andζ(4)=π4/90.
References[1]SmarandacheF.
Onlyproblems,notSolutions.
Chicago:XiquanPubl.
House,1993.
[2]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
[3]ZhangWenpeng.
ResearchonSmarandacheproblemsinnumbertheory.
HexisPublishingHouse,2004.
AFORMULAFORSMARANDACHELCMRATIOSEQUENCEWangTingDepartmentofMathematics,Xi'anScienceandArtsCollege,Xi'an,Shaanxi,P.
R.
ChinaAbstractInthispaper,areductionformulaforSmarandacheLCMratiosequencesSLR(5)isgiven.
Keywords:SmarandacheLCMratiosequences;Reductionformula.
§1.
IntroductionLet(x1,x2,.
.
.
,xt)and[x1,x2,.
.
.
,xt]denotethegreatestcommondivisorandtheleastcommonmultipleofanypostiveintegersx1,x2,.
.
.
,xtrespec-tively.
Letrbeapositiveintegerwithr>1.
Foranypositiveintegern,letT(r,n)=[n,n+1,.
.
.
,n+r1][1,2,.
.
.
r],thenthesequencesSLR(r)={T(r,n)}∞iscalledSmarandacheLCMra-tiosequencesofdegreer.
Inreference[1],MaohuaLestudiedthepropertiesofSLR(r),andgavetworeductionformulasforSLR(3)andSLR(4).
Inthispaper,westudythecalculatingproblemofSLR(5),andprovethefollowingconclusion.
Theorem.
Foranypostiveintegern,wehavethecalculatingformula:T(5,n)=11440n(n+1)(n+2)(n+3)(n+4),ifn≡0,8mod12;1120n(n+1)(n+2)(n+3)(n+4),ifn≡1,7mod12;1720n(n+1)(n+2)(n+3)(n+4),ifn≡2,6mod12;1360n(n+1)(n+2)(n+3)(n+4),ifn≡3,5,9,11mod12;1480n(n+1)(n+2)(n+3)(n+4),ifn≡4mod12;1240n(n+1)(n+2)(n+3)(n+4),ifn≡10mod12.
§2.
ProofofthetheoremTocompletetheproofofTheorem,weneedfollowingseveralsimpleLem-mas.
Lemma1.
Foranypostiveintegeraandb,wehave(a,b)[a,b]=ab.
46RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIILemma2.
Foranypostiveintegerswiths1andanyxedpositiveintegersmandk,wehavetheasymptoticformulan≤x1δk(bm(n))=x22ζ(m)pkpm(pmpm1+1)(p2m1)p|kpm+1(p+1)(pm1)+Ox32+,ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(10271093)48RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIwhereδk(n)denedasfollowing:δk(n)=max{d∈N|d|n,(d,k)=1},pdenotestheproductoverallprimenumbers.
Fromthistheorem,wemayimmediatelygetthefollowing:Corollary1.
Leta(n)bethesquarecomplementssequence,thenforanyrealnumberx>1andanyxedpositiveintegerk,wehavetheasymptoticformulan≤x1δk(a(n))=3x2π2pkp4p3+p2(p41)p|kp3(p+1)(p21)+Ox32+.
Corollary2.
Letb(n)bethecubiccomplementssequence,thenforanyrealnumberx>1andanyxedpositiveintegerk,wehavetheasymptoticformulan≤x1δk(b(n))=x22ζ(3)pkp6p5+p3(p61)p|kp4(p+1)(p31)+Ox32+.
§2.
ProofofthetheoremInthissection,wewillcompletetheproofofTheorem.
Lets=σ+itbeacomplexnumberandf(s)=∞n=11δk(bm(n))ns.
Notethat1δk(bm(n))1,soitisclearthatf(s)isaDirichletseriesabsolutelyconvergentforRe(s)>1.
BytheEulerproductformula[2]andthemultiplicativepropertyofδk(bm(n))wegetf(s)=∞n=11δk(bm(n))ns(1)=p1+1δk(bm(p))ps+1δk(bm(p2))p2s1δk(bm(pm))pms+···=p1+1δk(pm1)ps1δk(1)pms+1δk(pm1)p(m+1)s1δk(1)p2ms+···=pk1+pm(s1)1(ps11)(pms1)p|k111ps=ζ(s1)ζ(m(s1))pk(ps11)p(m1)(s1)pm(s1)1+p(m1)(s1)(pms1)Onthem-thpowercomplementssequence149*p|kps(pm(s1)p(m1)(s1))(ps1)(pm(s1)1),(2)whereζ(s)istheRiemannzeta-functionandpdenotestheproductoverallprimenumbers.
From(1)andPerron'sformula[3],wehaven≤x1δk(bm(n))(3)=12πi52+iT52iTζ(s1)ζ(m(s1))pk(ps11)p(m1)(s1)(pm(s1)1)+p(m1)(s1)(pms1)*p|kps(pm(s1)p(m1)(s1))(ps1)(pm(s1)1)·xssds+Ox32+T,(4)whereisanyxedpositivenumber.
Nowwemovetheintegrallinein(2)froms=52±iTtos=32±iT.
Thistime,thefunctionζ(s1)ζ(m(s1))pk(ps11)p(m1)(s1)(pm(s1)1)+p(m1)(s1)(pms1)*p|kps(pm(s1)p(m1)(s1))(ps1)(pm(s1)1)·xsshasasimplepolepointats=2withresiduex22ζ(m)pkpm(pmpm1+1)(p2m1)p|kpm+1(p+1)(pm1).
(5)Hence,wehave12πi52iT32iT+52+iT52iT+32+iT52+iT+32iT32+iTζ(s1)ζ(m(s1))*pk(ps11)p(m1)(s1)(pm(s1)1)+p(m1)(s1)(pms1)(6)*p|kps(pm(s1)p(m1)(s1))(ps1)(pm(s1)1)·xssds=x22ζ(m)pkpm(pmpm1+1)(p2m1)p|kpm+1(p+1)(pm1).
(7)50RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIWecaneasilygettheestimate12πi52iT32iTf(s)·xssdsx52+T,(8)12πi32+iT52+iTf(s)·xssdsx52+T,(9)and12πi32iT32+iTf(s)·xssdsx32+.
(10)TakingT=x,combining(2),(4),(5),(6)and(7)wemaydeducethatn≤x1δk(bm(n))=x22ζ(m)pkpm(pmpm1+1)(p2m1)p|kpm+1(p+1)(pm1)+Ox32+.
ThiscompletestheproofofTheorem.
References[1]SmarandacheF.
Onlyproblems,notSolutions.
Chicago:XiquanPubl.
House,1993.
[2]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
[3]PanChengdongandPanChengbiao.
FoundationofAnalyticNumberTheory.
Beijing:SciencePress,1997.
[4]ZhuWeiyi.
Onthek-powercomplementsandk-powerfreenumbersequence.
SmarandacheNotionsJournal,2004,14:66-69.
[5]LiuHongyanandLouYuanbing.
Anoteonthe29-thSmarandache'sproblem.
SmarandacheNotionsJournal,2004,14:156-158.
ONTHESMARANDACHECEILFUNCTIONANDTHEDIRICHLETDIVISORFUNCTIONRenDongmeiResearchCenterforBasicScience,Xi'anJiaotongUniversity,Xi'an,Shaanxi,P.
R.
ChinaAbstractThemainpurposeofthispaperisusingtheelementarymethodstostudythemeanvaluepropertiesofthecompositefunctioninvolvingDirichletdivisorfunc-tionandSmarandacheceilfunction,andgiveaninterestingasymptoticformulaforit.
Keywords:SmarandacheCeilfunction;Dirichletdivisorfunction;Meanvalue;Asymptoticformula.
§1.
IntroductionForaxedpositiveintegerkandanypositiveintegern,theSmarandacheceilfunctionSk(n)isdenedasfollowing:Sk(n)=min{m∈N:n|mk}.
ThisfunctionwasintroducedbyProfessorSmarandeche(seereference[1]).
Aboutthisfunction,manyscholarsstudieditsproperties(seereference[2]and[3]).
Inreference[2],Ibstedtpresentedthefollowingproperty:(a,b∈N)(a,b)=1=Sk(ab)=Sk(a)Sk(b).
Thatis,Sk(n)isamultiplicativefunction.
Inthispaper,weuseelementarymethodstostudythemeanvaluepropertiesofthecompositefunctioninvolvingd(n)andSk(n),andgiveaninterestingasymptoticformulaforit.
Thatis,weshallprovethefollowing:Theorem.
Letkbeagivenpositiveintegerwithk≥2.
Thenforanyrealnumberx≥1,wehavetheasymptoticformula:n≤xd(Sk(n))=6ζ(k)xlnxπ2p11pk+pk1+Cx+O(x12+ε).
whereCisacomputableconstant,andεisanyxedpositivenumber.
Takingk=2andk=4inTheorem1,notingthat:ζ(2)=π26,ζ(4)=π490,52RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIwemayimmediatelydeducethefollowing:Corollary.
Foranyrealnumberx≥1,wehavetheasymptoticformula:n≤xd(S2(n))=xlnxp11p2+p+C1x+O(x12+ε),n≤xd(S4(n))=π2xlnx15p11p4+p3+C2x+O(x12+ε),whereC1,C2arecomputableconstants.
Ifk=1,thenfunctionSk(n)turnsintotheidenticaltransformationandthecompositefunctiond(Sk(n))turnsintod(n).
Wehavethefollowingasymp-toticformula:n≤xd(S1(n))=n≤xd(n)=xlnx+(2γ1)x+O(x12),whereγistheEuler'sConstant.
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofthetheorem.
Letf(s)=∞n=1d(Sk(n))ns.
FromtheEulerproductformula(seereference[5])andthemultiplicativeprop-ertyofSk(n),wehavef(s)=p1+d(Sk(p))ps+d(Sk(p2))p2s+···=p1+d(p)psd(p)pks+d(p2)p(k+1)sd(p2)p2ks+···=p1+2ps2pks+3p(k+1)s+···=p1+11pks11ps*2ps+3p(k+1)s+···=ζ(s)p1+1ps*1+1pks+1p2ks+···=ζ(s)ζ(ks)p11pks+1ps=ζ(s)ζ(ks)ζ(s)ζ(2s)p11pks+p(k1)s,OntheSmarandacheceilfunctionandtheDirichletdivisorfunction53whereζ(s)istheRiemannzeta-function.
Itisobviousthat|d(Sk(n))|≤n,∞n=1d(Sk(n))nσ≤1σ1,whereσ>1istherealpartofs.
ByPerronformula(seereference[4]),lets0=0,b=32,T=x,thenwehaven≤xd(Sk(n))=12πi32+iT32iTζ2(s)ζ(ks)ζ(2s)R(s)xssds+O(x12+ε),whereR(s)=p11pks+p(k1)s,andεisanyxedpositivenumber.
Nowwecalculatetheterm12πi32+iT32iTζ2(s)ζ(ks)ζ(2s)R(s)xssds.
Wemovetheintegrallinefrom32±iTto12±iT.
Thenthefunctionζ2(s)ζ(ks)ζ(2s)R(s)xsshasasecondorderpoleats=1withresiduelims→1(s1)2ζ2(s)ζ(ks)ζ(2s)R(s)xss=lims→1(s1)2ζ2(s)ζ(ks)ζ(2s)R(s)xss+(s1)2ζ2(s)ζ(ks)ζ(2s)R(s)sxslnxxss2=6ζ(k)xln(x)π2p11pk+pk1+Cx,whereCisacomputableconstant.
Sowecanobtain12πi32+iT32iT+12+iT32+iT+12iT12+iT+32iT12iTζ2(s)ζ(ks)ζ(2s)R(s)xssds=6ζ(k)xln(x)π2p11pk+pk1+Cx.
TakingT=x,wehavetheestimate12πi12+iT32+iT+32iT12iTζ2(s)ζ(ks)ζ(2s)R(s)xssds1,wehavetheasymptoticformulan≤xd(S1(n))=xlnx+(2γ1)x+Ox13andn≤xd(Sk(n))=ζ(k)x+ζ1kx1k+Ox1k+1,56RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIwhereγistheEulerconstant,andζ(n)istheRiemannzeta-function.
FromthisTheoremwemayimmediatelydeducethefollowingCorollary1.
Foranyrealnumberx>1,wehavetheasymptoticformulan≤xd(S2(n))=π26x+ζ12x12+Ox13.
Corollary2.
Foranyrealnumberx>1,wehavetheasymptoticformulan≤xd(S4(n))=π490x+ζ14x14+Ox15.
§2.
ProofofthetheoremInthissection,wewillgivetheproofofthetheorem.
ThefollowingLemmaisnecessary.
Lemma.
Ifx≥1ands≥0,s=1,wehaven≤x1ns=x1s1s+ζ(s)+O(xs),whereζ(s)=1+1s1s∞1t[t]ts+1dt.
Proof.
ThisLemmacanbeeasilyprovedbyusingtheEulersummationfor-mula,seeTheorem3.
2(b)of[3].
Nowwecometotheproofofthetheorem.
ItisobviousthatS1(n)=n,andthisdeducestherstpartofthetheoremimmediatelybytheclassicalresultonDirichletdivisorproblem,seereference[4].
Nowassumek≥2,wehaven≤xd(Sk(n))=n≤xd|Sk(n)1,fromthedenitionofSk(n)weknowthatd|Sk(n)dk|n,hencen≤xd(Sk(n))=n≤xdk|n1=dkl≤x1.
Letδ=x1k+1,applyingtheaboveformulaandTheorem3.
17of[3]wehaven≤xd(Sk(n))=dkl≤x1OnadualfunctionoftheSmarandacheceilfunction57=1≤dk≤δk1≤l≤x/dk1+1≤l≤δ1≤dk≤x/l11≤dk≤δk11≤l≤δ1=1≤d≤δ1≤l≤x/dk1+1≤l≤δ1≤dk≤x/l1[δ]2=1≤d≤δxdk+1≤l≤δxl1/k(δ22δ{δ}+{δ}2).
UsingtheaboveLemmawegetn≤xd(Sk(n))=x1≤d≤δ1dk+x1/k1≤l≤δ1l1/k+O(δ)(δ2+O(δ))=xδ1k1k+ζ(k)+O(δk)+x1/kδ11k11k+ζ1k+O(δ1k)δ2+O(δ).
Noticethatx=δk+1,wehaven≤xd(Sk(n))=δ21k+ζ(k)x+δ211k+ζ1kx1kδ2+O(δ)=ζ(k)x+ζ1kx1k+O(δ)=ζ(k)x+ζ1kx1k+O(x1k+1).
ThiscompletestheproofofTheorem.
References[1]F.
Smarandache.
OnlyProblems,Notsolutions.
Chicago:XiquanPub-lishingHouse,1993.
[2]SabinTabircaandTatianaTairca.
SomenewresultsconcerningtheSmarandacheceilfunction,SmarandacheNotions,bookseriesVol.
13,2002,30-36.
[3]TomM,Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag.
1976[4]PanChengdongandPanChengbiao.
FoundationofAnalyticNumberTheory.
Beijing:SciencePress,1999.
ONTHELARGESTM-THPOWERNOTEXCEEDINGNLiuDuansenandLiJunzhuangInstituteofMathematics,ShangluoTeacher'sCollege,Shangluo,Shaanxi,P.
R.
ChinaAbstractInthispaper,someelementarymethodsareusedtostudythepropertiesofthelargestm-thpowernotexceedingn,andgiveanidentityaboutit.
Keywords:Thelargestm-thpowernotexcedingn;Dirichletseries;Identity.
§1.
IntroductionandresultsLetnisapositiveinteger.
Itisclearthatthereexistsanintegerksuchthatkm≤n1,theDirichletseriesf(s)isconvergentandf(s)=C1mζ(msm+1)+C2mζ(msm+2)+···+Cm1mζ(ms1)+ζ(ms),whereCnm=m!
n!
(mn)!
,andζ(s)istheRiemannzeta-function.
Fromthistheoremwemayimmediatelydeducethefollowing:ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(10271093)andtheEducationDepartmentFoundationofShannxiProvince(04JK132)60RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIICorollary1.
Takingm=2ands=3/2orm=s=2intheaboveTheorem,thenwehavetheidentities∞n=11b232(n)=π23+ζ(3)and∞n=11b22(n)=2ζ(3)+π490.
Corollary2.
Takingm=3ands=2orm=2ands=3intheabovetheorem,wehavetheidentities∞n=11b32(n)=π430+3ζ(5)+π6945and∞n=11b23(n)=2ζ(5)+π6945.
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofthetheorem.
Foranypositiveintegern,letbm(n)=km.
Itisclearthatthereareexactly(k+1)mkmintegernsuchthatbm(n)=km.
Sowemaygetf(s)=∞n=11bsm(n)=∞k=1(k+1)mkmkms=∞k=1C1mkm1+C2mkm2Cm1mk+1kms.
Fromtheintegralcriterion,weknowthatf(s)isconvergentifms(m1)>1.
Thatis,s>1.
Ifs>1,notethatζ(s)=∞n=11ns,wehavef(s)=C1mζ(msm+1)+C2mζ(msm+2)Cm1mζ(ms1)+ζ(ms).
ThiscompletestheproofofTheorem.
It'seasytocomputethatf(2)=C1mζ(m+1)+C2mζ(m+2)Cm1mζ(2m1)+ζ(2m),f(3)=C1mζ(2m+1)+C2mζ(2m+2)Cm1mζ(3m1)+ζ(3m).
Nowthecorollariesfollowsfromζ(2)=π2/6,ζ(4)=π4/90,ζ(6)=π6/945(seereference[4])andtheabovetheorem.
References[1]F.
Smarandache.
OnlyProblems,Notsolutions.
Chicago:XiquanPub-lishingHouse,1993.
Onthelargestm-thpowernotexceedingn161[2]ZhangWenpeng.
Onthecubepartsequenceofapositiveinteger.
JournalofXianYangTeacher'sCollege,2003,18:5-7.
[3]ZhengJianfeng.
Ontheinferiorandsuperiork-thpowerpartofapos-itiveintegeranddivisorfunction.
SmarandacheNotionsJournal,2004,14:88-91.
[4]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
ONTHESMARANDACHEBACKCONCATENATEDODDSEQUENCESLiJunzhuangandWangNianliangInstituteofMathematics,ShangluoTeacher'sCollegeShangluo,Shaanxi,P.
R.
ChinaAbstractThemainpurposeofthispaperistostudythearithmeticalpropertiesoftheSmarandachebackconcatenatedoddsequences,andgiveseveralsimpleproper-tiesinvolvingtherecursionformulaandexactexpressionsforthegeneralterm.
Keywords:TheSmarandachebackconcatenatedoddsequence;Arithmeticalproperties;Re-cursionformula.
§1.
IntroductionandmainresultsThefamousSmarandachebackconcatenatedoddsequence{bn}isdenedasfollowing1,31,531,7531,97531,1197531,131197531,15131197531,1715131197531,Inproblem3ofreference[1],ProfessorMihalyBenczeandLucianTutescuaskedustostudythearithmeticalpropertiesaboutthisse-quence.
Itisinterestingforustostudythisproblem.
Butit'sapitythatnonehadstudieditbefore.
Atleastwehaven'tseensuchapaperyet.
Inthispaper,weshallusetheelementarymethodstostudythearithmeticalpropertiesoftheSmarandachebackconcatenatedoddsequences,andgiveseveralsimpleprop-ertiesinvolvingtherecursionformula,exactexpressionsforthegeneralterm,andsoon.
Thatis,weshallprovethefollowing:Theorem1.
Letn≥2beanypositiveintegerwith(2n3)haskdigits.
ThenfortheSmarandachebackconcatenatedoddsequences{bn},wehavethefollowingrecursionformulabn=bn1+(2n1)*101010k18+k*(n1),whereb1=1.
ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(10271093)andtheEducationDepartmentFoundationofShannxiProvince(04JK132)64RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIITheorem2.
Let(2n3)haskdigits.
Thenthebn1-thtermintheSmarandachebackconcatenatedoddsequences{bn}has1010k18+k*(n1)digits.
Theorem3.
LetSk(m,n)=ni=m(2i1)*10k(i1),thenwehavethefollowingexactexpressionforthegeneralterm.
Thatis,bn=1+S1(2,6)+105*S2(7,51)+1055*S3(52,501)+···+10(k1)5···5*Sk(m,n).
Theorem4.
LetS50denotesthesummationfortherst50termsintheSmarandachebackconcatenatedoddsequences{bn},thenwehaveS50=11*45*106201*509+81*106+97092+4*106+4*10093+99*109513*44*10799+95*1095+73*109992+4*1095+4*1011993.
§2.
SomelemmasLemma1.
Letk,m,narepositiveintegerswithm≤n,thenwehaveSk(m,n)=(2m1)*10k(m1)110k+2*10km[110k(nm)](110k)2+(2n1)*10kn110k.
Proof.
FromthedenitionofSk(m,n),wehaveSk(m,n)=ni=m(2i1)*10k(i1)=(2m1)*10k(m1)+(2m+1)*10km2n1)*10k(n1)and10kSk(m,n)=(2m1)*10km+(2m+1)*10k(m+1)+···+(2n1)*10kn.
Thuswecanget(110k)Sk(m,n)=(2m1)*10k(m1)+2*[10km+10k(m+1)OntheSmarandachebackconcatenatedoddsequences16510k(n1)](2n1)*10kn=(2m1)*10k(m1)+2*10km[110k(nm)]110k(2n1)*10kn.
SowehaveSk(m,n)=(2m1)*10k(m1)110k+2*10km[110k(nm)](110k)2+(2n1)*10kn110k.
ThisprovesLemma1.
Lemma2.
LetSk(n,m)=ni=mi2*10ki.
Thenforanypositiveintegersk,m,n,wehaveSk(m,n)=m2*10km110k+(2m+1)*10k*(m+1)(110k)2+2*10k(m+2)[110k(nm1)](110k)3(2n1)*10k*(n+1)(110k)2n2*10k(n+1)110k.
Proof.
FromthedenitionofSk(m,n),wehaveSk(m,n)=m2*10k*m+(m+1)2*10k*(m+1)+(m+2)2*10k*(m+2)m+(nm)]2*10k[m+(nm)]and10kSk(m,n)=m2*10k*(m+1)+(m+1)2*10k*(m+2)m+(nm)]2*10k[m+(nm)+1].
Thuswecanget(110k)Sk(m,n)=m2*10k*m+(2m+1)*10k*(m+1)+(2m+3)*10k*(m+2)2[m+(nm)]1}*10k[m+(nm)][m+(nm)]2*10k[m+(nm)]+1.
66RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIISowehaveSk(m,n)=m2*10km110k+ni=m+1(2i1)*10k*i110k+n2*10k(n+1)110k=m2*10km110k+(2m+1)*10k*(m+1)(110k)2+2*10k(m+2)[110k(nm1)](110k)3(2n1)*10k*(n+1)(110k)2n2*10k(n+1)110k.
ThisprovesLemma2.
Lemma3.
LetSk(n,m)=ni=mi*10ki.
Thenforanypositiveintegersk,m,n,wehaveSk(m,n)=m2*10km110k+10k(m+1)[110k(nm)](110k)2n*10k(n+1)(110k).
Proof.
FromthedenitionofSk(m,n),wehaveSk(m,n)=m*10km+(m+1)*10k(m+1)m+(nm)]*10k[m+(nm)]and10kSk(m,n)=m*10k(m+1)+(m+1)*10k(m+2)m+(nm)]*10k[m+(nm)+1].
Thuswecanget(110k)Sk(m,n)=m*10km+10k(m+1)10k[m+(nm)][m+(nm)]*10k[m+(nm)+1].
SowehaveSk(m,n)=m*10km110k+*10k(m+1)[110k(nm)](110k)2n*10k(n+1)(110k).
ThisprovesLemma3.
OntheSmarandachebackconcatenatedoddsequences267§3.
ProofofthetheoremsInthissection,weshallusetheabovelemmastocompletetheproofofthetheorems.
FirstweusemathematicalinductiontoproveTheorem1.
(i)Ifn=2,3,itisobviousthatTheorem1istrue.
(ii)AssumingthatTheorem1istrueforn=m.
Thatis,thefollowingrecursionformulabm=bm1+(2m1)*101010k18+k*(m1)istruefortheSmarandachebackconcatenatedoddsequence{bn}.
Comparingthedifferencebetweenbmandbm1,weknowthatbm1has1010k18+k*(m1)digits.
Ifn=m+1,wecandivideitintotwocases:(i)If2m1stillhaskdigits,thenweknowthatbmhas[1010k18+k*(m1)+k]digits.
Comparingthedifferencebetweenbm+1andbm,wemayimmediatelydeducethatbm+1bm2m+1=101010k18+k*(m1)+k.
Thatis,bm+1=bm+(2m+1)*101010k18+k*m.
(ii)If2m1hask+1digits,thenweknowthatbmhas[1010k18+k*(m1)+k+1]digits.
Recallthat(2m3)haskdigitsand(2m1)hask+1digits,whichexistonlyinthecasethat2m3=10k1,2m1=10k+1.
Thatis,m=10k2+1.
Therefore,wehave1010k18+k*(m1)+(k+1)=1010k18+k*10k2+(k+1)=1010k+118+(k+1)*mComparingthedifferencebetweenbm+1andbm,wecandeducethatbm+1bm2m+1=101010k18+k*(m1)+(k+1)=101010k+118+(k+1)*m.
Thatis,bm+1=bm+(2m+1)*101010k+118+(k+1)*m.
Combining(i)and(ii),Theorem1istrueforanypositiveintegern.
ThiscompletestheproofofTheorem1.
ByusingtheresultofTheorem1,andnotethatthedifferencebetweenbnandbn1,wecanimmediatelygettheresultofTheorem2.
68RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIByTheorem1,when(2n3)haskdigits,let(2m3)betheleastpositiveintegerwhichhaskodddigits(herem=2whenk=1,andm=10k12+2).
Thenwehavebn=1+(2*21)*101010118+1*(21)+(2*31)*101010118+1*(31)+(2*41)*101010118+1*(41)+(2*51)*101010118+1*(51)+(2*61)*101010118+1*(61)+(2*71)*101010218+2*(71)+(2*81)*101010218+2*(81)2*511)*101010218+2*(511)+(2*521)*101010318+3*(521)+(2*531)*101010318+3*(531)2*m1)*101010k18+k*(m1)2*n1)*101010k18+k*(n1)+1+101010118*S1(2,6)+101010218*S2(7,51)+101010318*S3(52,501)101010k18*Sk(m,n)+1+101010118*S1(2,6)+101010218*S2(7,51)+101010318*S3(52,501)101010k18*Sk(m,n)=1+S1(2,6)+105*S2(7,51)+1055*S3(52,501)10(k1)5···5*Sk(m,n).
ApplyingtheresultofLemma1,wecandeduceTheorem3.
NowwecometoproveTheorem4.
ByusingtheresultofTheorem1,wehaveS50=b1+b2+b3b49+b50=50+49*(2*21)*101010118+1*(21)+48*(2*31)*101010118+1*(31)+47*(2*41)*101010118+1*(41)+46*(2*51)*101010118+1*(51)+45*(2*61)*101010118+1*(61)OntheSmarandachebackconcatenatedoddsequences369+44*(2*71)*101010218+2*(71)+43*(2*81)*101010218+2*(81)3*(2*481)*101010218+2*(481)+2*(2*491)*101010218+2*(491)+1*(2*501)*101010218+2*(501)=50*[2*(5150)1]*101010118+1*(5050)+49*[2*(5149)1]*101010118+1*(5049)+48*[2*(5148)1]*101010118+1*(5048)+47*[2*(5047)1]*101010118+1*(5047)45*[2*(5145)1]*101010118+1*(5145)+44*[2*(5044)1]*101010218+2*(5044)2*[2*(512)1]*101010218+2*(502)+1*[2*(511)1]*101010218+2*(501)=101*[50*101010118+1*501*50+49*101010118+1*501*4945*101010118+1*501*45+44*101010218+2*502*442*101010218+2*502*2+1*101010218+2*502*1]2*[502*101010118+1*501*50+492*101010118+1*501*49452*101010118+1*501*45+442*101010218+2*502*4422*101010218+2*502*2+12*101010218+2*502*1]=101*50i=45i*101010118+1*501*i+101*44i=1i*101010218+2*502*i70RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII2*50i=45i2*101010118+1*501*i2*44i=1i2*101010218+2*502*i=101*1050*50i=45i*10i+101*1095*44i=1i*102*i2*1050*50i=45i2*10i2*1095*44i=1i2*102*i.
ApplyingtheresultsofLemma2andLemma3,wecangetS50=11*45*106201*509+81*106+97092+4*106+4*10093+99*109513*44*10799+95*1095+73*109992+4*1095+4*1011993.
ThiscompletestheproofofTheorem4.
References[1]MihalyBenczeandLucianTutescu(Eds.
).
Somenotionsandquestionsinnumbertheory.
Vol2.
http://www.
gallup.
unm.
edu/smarandache/SNAQINT2.
TXT.
[2]F.
Smarandache,Onlyproblems,notSolutions,XiquanPubl.
House,Chicago,1993.
[3]TomM.
Apostol,IntroductiontoAnalyticNumberTheory,Springer-Verlag,NewYork,HeidelbergBerlin,1976.
ONTHEADDITIVEHEXAGONNUMBERSCOMPLEMENTSLichaoandYangCundianInstituteofMathematics,ShangluoTeacher'sCollege,Shangluo,Shaanxi,P.
R.
ChinaAbstractInthispaper,similartotheSmarandachek-thpowercomplements,wedenedthehexagonnumberscomplements.
Usingtheelementarymethod,westudiedthemeanvaluepropertiesoftheadditivehexagonnumberscomplements,andobtainedsomeinterestingasymptoticformulaeforit.
Keywords:Additivehexagonnumberscomplements;Meanvalue;Asymptoticformula.
§1.
IntroductionandresultsLetnbeapositiveinteger.
Ifthereexistsapositiveintegermsuchthatn=m(2m1),thenwecallnasahexagonnumber.
Foranypositiveintegern,theSmarandachek-thpowercomplementsbk(n)isthesmallestpositivein-tegersuchthatnbk(n)iscompletek-thpower,seeproblem29of[1].
SimilartotheSmarandachek-thpowercomplements,wedenetheadditivehexagonnumberscomplementsa(n)asfollows:a(n)isthesmallestnonnegativeinte-gersuchthata(n)+nisahexagonnumber.
Forexample,ifn=1,2,···15,wehavetheadditivehexagonnumbersequences{a(n)}(n=1,2,···15)asfollows:0,4,3,2,1,0,8,7,6,5,4,3,2,1,0.
Inthispaper,westudythemeanvaluepropertiesofthecompositearithmeticfunctiond(a(n))(whered(n)istheDirichletdivisorfunction),andgivesomeinterestingasymptoticformulaeforit.
Thatis,weshallprovethefollowingconclusions:Theorem1.
Foranyrealnumberx≥3,wehavetheasymptoticformula:n≤xa(n)=2√23x32+O(x).
Theorem2.
Foranyrealnumberx≥3,wehavetheasymptoticformula:n≤xd(a(n))=12xlogx+32log2+(2γ1)12x+O(x23),ThisworkissupportedbyN.
S.
F.
ofP.
R.
China(10271093)andtheEducationDepartmentFoundationofShannxiProvince(04JK132)72RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIwhereγistheEulerconstant.
§2.
SomeLemmasBeforetheproofofthetheorems,someLemmaswillbeuseful.
Lemma1.
Foranyrealnumberx≥3,wehavetheasymptoticformula:n≤xd(x)=xlnx+(2γ1)x+Ox13,whereγistheEulerconstant.
Proof.
Seereference[2].
Lemma2.
Foranyrealnumberx≥3andanynonnegativearithmeticalfunctionf(n)withf(0)=0,wehavetheasymptoticformula:n≤xf(a(n))=x122m=1i≤4mf(i)+Oi≤4x122f(i),where[x]denotesthegreatestintegerlessthanorequaltox.
Proof.
Foranyrealnumberx≥1,letMbeaxedpositiveintegersuchthatM(2M1)≤x√n,thenS(n)=p(n).
Proof.
Ifnbeasquare-freenumber,letn=p1p2···prp(n),thenpi√n,thenp2(n)n.
Letn=pα11pα22···pαrrp(n),sowehavepα11pα22···pαrr√n}.
NotethatS(n)p(n)lnnandσ(n)nlnlnn,wehavetheestimaten∈Aσ(S(n))n∈AS(n)lnln(S(n))n≤x√nlnnlnlnnx32lnxlnlnx.
(1)ThenusingtheaboveLemmaswemayimmediatelygetn∈Bσ(S(n))=n≤xp(n)>√nσ(p(n))=n≤√x√n≤p≤xnσ(p)=n≤√x√x≤p≤xn(p+1)+On≤√x√n≤p≤√xp=n≤ln2xx22n2lnxn+ln2x≤n≤√xx22n2lnxn+O(x32lnx)=π212x2lnx+Ox2ln2x.
(2)Combining(1)and(2)weobtainn≤xσ(S(n))=n∈Aσ(S(n))+n∈Bσ(S(n))=π212x2lnx+Ox2ln2x.
ThiscompletestheproofofTheorem.
References[1]F.
Smarandache.
Onlyproblems,Notsolutions.
Chicago:XiquanPubl.
House,1993.
[2]LiHailongandZhaoXiaopeng.
ResearchonSmarandacheProblemsInNumberTheory.
Hexis,2004,119-122.
[3]MarkFarrisandPatrickMitchell.
BoundingtheSmarandachefunction.
SmarandacheNotionsJournal,Vol13(2002),37-42.
[4]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
Springer-Verlag:NewYork,1976.
ONTHEMEANVALUEOFANEWARITHMETICALFUNCTIONZhaoJianandLiChaoInstituteofMathematics,ShangluoTeacher'sCollege,Shangluo,Shaanxi,P.
R.
Chinazzz@zzz.
comAbstractThemainpurposeofthispaperisusingtheelementarymethodstostudythecon-vergentpropertiesofanewDirichlet'sseriesinvolvingthetriangularnumbers,andgiveaninterestingidentityforit.
Keywords:Dirichlet'sseries;Convergence;Identity.
§1.
IntroductionandresultsForanypositiveintegerm,itisclearthatm(m+1)/2isapositiveinte-ger,andwecallitasthetriangularnumber,becausetherearecloserelationsbetweenthesenumbersandthegeometry.
Nowforanypositiveintegern,wedenearithmeticalfunctionc(n)asfollows:c(n)=max{m(m+1)/2:m(m+1)/2≤n,m∈N}.
Thatis,c(n)isthegreatesttriangularnumber≤n.
Fromthedenitionofc(n)wecaneasilydeducethatc(1)=1,c(2)=1,c(3)=3,c(4)=3,c(5)=3,c(6)=6,c(7)=6,c(8)=6,c(9)=6,c(10)=10,Aboutthisfunction,itseemsthatnonehadstudieditbefore,evenwedonotknowitsarithmeticalproperties.
Inthispaper,weintroduceanewDirichlet'sseriesf(s)involvingthesequences{c(n)},i.
e.
,f(s)=∞n=11cs(n).
Thenweusingtheelementarymethodstostudytheconvergentpropertiesoff(s),andobtainaninterestingidentity.
Thatis,weshallprovethefollowingresult:TheoremLetsbeanypositiverealnumber.
ThentheDirichlet'sseriesf(s)isconvergentifandonlyifs>1.
Especiallyfors=2,3and4,wehaveThisworkissupportedbyN.
S.
F.
ofP.
R.
China(10271093)andtheEducationDepartmentFoundationofShannxiProvince(04JK132)80RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIItheidentities:∞n=11c2(n)=2π234;∞n=11c3(n)=8ζ(3)4π2+32;∞n=11c4(n)=π4548ζ(3)+1603π2384;6f(3)+f(4)=π45+883π2192.
whereζ(k)istheRiemannzeta-function.
§2.
ProofofthetheoremInthissection,wewillcompletetheproofofthetheorem.
Itiseasytoseethatifm(m+1)2≤n1,divergentifs≤1.
Speciallyifs=2,wecanwritef(2)=4∞m=11m2(m+1)=4∞m=11m21m+1m+1=4ζ(2)4.
Usingthesamemethod,wecanalsoobtainf(3)=8ζ(3)24ζ(2)+32;f(4)=16ζ(4)48ζ(3)+320ζ(2)384.
Nowthetheoremfollowsfromtheidentities(seereference[2])ζ(2)=π2/6andζ(4)=π4/90.
Thiscompletetheproofofthetheorem.
Onthemeanvalueofanewarithmeticalfunction181Note:Infact,foranypositiveintegers≥2,usingourmethodswecanexpressf(s)astheRiemannzeta-function.
References[1]SmarandacheF.
Onlyproblems,notSolutions.
Chicago:XiquanPubl.
House,1993.
[2]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
ONTHESECONDCLASSPSEUDO-MULTIPLESOF5SEQUENCESGaoNanSchoolofScience,Xi'anShiyouUniversity,Xi'an,Shaanxi,P.
R.
ChinaAbstractThemainpurposeofthispaperistostudythemeanvaluepropertiesofthesecondclasspseudo-multiplesof5sequences,andgiveaninterestingasymptoticformulaforit.
Keywords:Secondclasspseudo-multiplesof5sequences;Meanvalue;Asymptoticfor-mula.
§1.
IntroductionandresultsApositiveintegeriscalledpseudo-multipleof5ifsomepermutationofitsdigitsisamultipleof5,includingtheidentitypermutation.
Inreference[1],ProfessorF.
Smarandacheaskedustostudythepropertiesofthepseudo-multipleof5sequences.
Aboutthisproblem,WangXiaoying[2]hadstudiedit,andobtainedsomeinterestingresults.
Now,wedenethesecondclasspseudo-multiplesof5numbersasfollowing:Apositiveintegeriscalledthesecondclasspseudo-multipleof5ifitisnotamultipleof5,butitssomepermutationofitsdigitsisamultipleof5.
Forconvenience,letAdenotesthesetofallpseudo-multipleof5sequences,andBdenotesthesetofallsecondclasspseudo-multipleof5sequences.
Inthispaper,weshallusetheelementarymethodstostudythemeanvaluepropertiesofthesecondclasspseudo-multipleof5sequences,andobtainaninterestingasymptoticformulaforit.
Thatis,weshallprovethefollowing:Theorem.
Foranyrealnumberx≥1,wehavetheasymptoticformulan∈Bn≤xd(n)=1625xlnx+2γ1+ln52+Oxln8ln10+ε,whered(n)istheDirichlet'sdivisorfunction,γistheEulerconstant,andεdenotesanyxedpositivenumber.
84RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII§2.
SomeLemmasTocompletetheproofoftheaboveTheorem,weneedthefollowingtwoLemmas:Lemma1.
Letqbeaxedprimeorq=1.
Thenforanyrealnumberx≥1,wehavetheasymptoticformulan≤xd(qn)=21qxlnx+2γ1+lnq2q1+Ox12+,whereγistheEulerconstant,anddenotesanyxedpositivenumber.
Proof.
Ifq=1,thenfromTheorem3.
3of[3]weknowthatLemma1iscorrect.
Nowforanyprimeqandrealnumbers>1,letf(s)=∞n=1d(qn)ns.
Notethatd(n)isamultiplicativefunction,sobytheEulerproductformula[3]wehavef(s)=∞n=1d(qn)ns=∞α=0∞n1=1(n1,q)=1d(qα+1n1)qsαns1=∞α=02+αqsα∞n1=1(n1,q)=1d(n1)ns1=111qs+111qs2*p(p,q)=11+d(p)ps+d(p2)p2sd(pk)pks+···=111qs+111qs2ζ2(s)11qs2=ζ2(s)21qs.
ThenfromthisidentityandthePerron'sformula[4]wemayimmediatelydeducetheasymptoticformulan≤xd(qn)=21qxlnx+2γ1+lnq2p1+Ox12+.
ThiscompletestheproofofLemma1.
Onthesecondclasspseudo-multiplesof5sequences85Lemma2.
Foranyrealnumberx≥1,wehavetheasymptoticformulan∈An≤xd(n)=xlnx+(2γ1)x+Oxln8ln10+ε.
Proof.
Foranyrealnumberx≥1,itisclearthatthereexistsanonnegativeintegerksuchthat10k≤x12,applyingLemma1withq=1wehaven∈An≤xd(n)=n≤xd(n)n∈An≤xd(n)=n≤xd(n)+On∈An≤xxε=n≤xd(n)+Oxln8ln10+ε=xlnx+(2γ1)x+Oxln8ln10+ε.
ThisprovesLemma2.
§3.
ProofofthetheoremInthissection,wecompletetheproofofTheorem.
FromthedenitionofsetAandsetB,weknowtherelationshipbetweenthem:AB={multiplesof5}.
CombiningtheaboveLemmaswemayimmediatelygetn∈Bn≤xd(n)=n∈An≤xd(n)5n≤xd(5n)=xlnx+(2γ1)x+Oxln8ln10+ε925xlnxln5+2γ1+ln59=1625xlnx+2γ1+ln52+Oxln8ln10+ε.
86RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIThiscompletestheproofofTheorem.
Notes:Infact,wemayusethesimilarmethodtostudyotherarithmeticalfunctionsonthepseudo-multiplesof5sequencesandthesecondclasspseudo-multiplesof5sequences.
References[1]F.
Smarandache.
Onlyproblems,notsolutions,XiquanPublishingHouse,Chicago,1993.
[2]WangXiaoying.
OntheSmarandachePseudo-multiplesof5Sequence,Hexis,2004,17-19.
[3]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory,NewYork,1976.
[4]PanChenddongandPanChengbiao.
ElementsoftheAnalyticNumberTheory,SciencePress,Beijing,1991.
ACLASSOFDIRICHLETSERIESANDITSIDENTITIESLouYuanbingCollegeofScience,TibetUniversity,Lhasa,Tibet,P.
R.
Chinazzz@zzz.
comAbstractInthispaper,weusingtheelementarymethodstostudytheconvergentprop-ertiesofoneclassDirichletseriesinvolvingaspecialsequences,andgiveaninterestingidentityforit.
Keywords:Convergentproperty;Dirichletseries;Identity.
§1.
IntroductionandresultsForanypositiveintegernandm≥2,wedenethem-thpowercomple-mentbm(n)isthesmallestpositiveintegersuchthatnbm(n)isacompletem-thpower,seeproblem29of[1].
Nowforanypositiveintegerk,wealsodeneanarithmeticfunctionδk(n)asfollows:δk(n)=max{d∈N|d|n,(d,k)=1},ifn=0,0,ifn=0.
LetAdenotesthesetofallpositiveintegersnsuchthattheequationδk(n)=bm(n).
Thatis,A={n:n∈N,δk(n)=bm(n)}.
Inthispaper,weusingtheelementarymethodstostudytheconvergentpropertiesoftheDirichletseriesinvolvingthesetA,andgiveaninterestingidentityforit.
Thatis,weshallprovethefollowingconclusion:Theorem.
Letmbeapositiveevennumber.
Thenforanyrealnumbers>1andpositiveintegerk,wehavetheidentity:∞n=1n∈A1ns=ζm2sζ(ms)p|kp32ms(pms1)p12ms1,whereζ(s)istheRiemannzeta-function,andpdenotestheproductoverallprimes.
FromthisTheoremwemayimmediatelydeducethefollowingcorollaries:88RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIICorollary1.
LetB={n:n∈N,δk(n)=b2(n)},thenwehave∞n=1n∈B1n2=15π2p|kp6(p41)(p21).
Corollary2.
LetC={n:n∈N,δk(n)=b4(n)},thenwehave∞n=1n∈C1n2=∞n=1n∈B1n4=105π4p|kp12(p81)(p41).
Corollary3.
LetC={n:n∈N,δk(n)=b4(n)},thenwehave∞n=1n∈C1n3=6756756911π6p|kp18(p121)(p61).
§2.
ProofofthetheoremInthissection,wewillcompletetheproofofthetheorem.
Foranyrealnumbers>0,itisclearthat∞n=1n∈A1ns1,thus∞n=1n∈A1nsisalsoconvergentifs>1.
NowwendthesetA.
Letn=pα11pα22···pαssdenotesthefactorizationofnintoprimepowers.
Firstfromthedenitionsofδk(n)andbm(n)wekonwthatδk(n)andbm(n)botharemultiplicativefunctions.
SoinordertondA,weonlydiscusstheproblemincasen=pα.
Ifn=pαand(p,k)=1,thenwehaveδk(pα)=pα;bm(pα)=pmα,if1≤α≤m;bm(pα)=pm+m[αm]α,ifα>mandα=m;bm(pα)=1,ifα=rm,whererisanypositiveinteger,and[x]denotesthegreatestinteger≤x.
Sointhiscaseδk(pα)=bm(pα)ifandonlyifα=m2.
Ifn=pαand(p,k)=1,thenδk(pα)=1,sointhiscasetheequationδk(pα)=bm(pα)hassolutionifandonlyifn=prm,r=0,1,2,NowfromtheEulerproductformula(see[2])andthedenitionofA,wehave∞n=1n∈A1ns=pk1+1pm2sp|k1+1pms+1p2ms+1p3ms+···AclassofDirichletseriesanditsidentities89=p1+1pm2sp|k111pmsp|k1+1pm2s1=ζm2sζ(ms)p|kp32ms(pms1)p12ms+1,whereζ(s)istheRiemannzeta-function,andpdenotestheproductoverallprimes.
ThiscompletestheproofofTheorem.
TheCorollariesfollowsfromζ(2)=π2/6,ζ(4)=π4/90,ζ(6)=π8/945,ζ(8)=π8/9450andζ(12)=691π8638512875.
References[1]F.
Smarandache.
Onlyproblems,notsolutions,XiquanPublishingHouse,Chicago,1993.
[2]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
ANARITHMETICALFUNCTIONANDTHEPERFECTK-THPOWERNUMBERSYangQianliandYangMingshunDepartmentofMathematics,WeinanTeacher'sCollege,Weinan,Shaanxi,P.
R.
ChinaAbstractForanyprimespandqwith(p,q)=1,thearithmeticalfunctionepq(n)denedasthelargestexponentofpowerpqwhichdividesn.
Inthispaper,weusetheelementarymethodstostudythemeanvaluepropertiesofepq(n)actingontheperfectk-thpowernumbersequences,andgiveaninterestingasymptoticformulaforit.
Keywords:theperfectk-thpowernumber;Asymptoticformula;Meanvalue.
§1.
IntroductionForanyprimep,letep(n)denotesthelargestexponentofpowerpwhichdividesn.
Inproblem68ofreference[1],ProfessorF.
Smarandacheaskedustostudythepropertiesofthisarithmeticalfunction.
Aboutthisproblem,manyscholarsshowedgreatinterestinit,andobtainedsomeinterestingresults,seereferences[2]and[3].
Similarly,wewilldenearithmeticalfunctionepq(n)asfollows:foranytwoprimespandqwith(p,q)=1,letepq(n)denotesthelargestexponentofpowerpqwhichdividesn.
Thatis,epq(n)=max{α:(pq)α|n,α∈N+}.
Accordingto[1],anumberniscalledaperfectk-thpowernumberifitsatisedk|αforallpαn,wherepαndenotespα|n,butpα+1n.
LetAdenotesthesetofalltheperfectk-thpowernumbers.
Itseemsthatnooneknowstherelationsbetweenthesetwoarithmeticalfunctionsbefore.
Themainpurposeofthispaperisusingtheelementarymethodstostudythemeanvaluepropertiesofepq(n)actingonthesetA,andgiveaninterestingasymptoticformulaforit.
Thatis,weshallprovethefollowing:Theorem.
Letpandqaretwoprimeswith(p,q)=1,thenforanyrealnumberx≥1,wehavetheasymptoticformulan≤xn∈Aepq(n)=Cp,qkx1k+x12k+,92RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIwhereCp,q=(p1)(q1)pq∞n=1n(pq)nisacomputablepositiveconstant,anddenotesanyxedpositivenumber.
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofthetheorem.
Firstwedenearithmeticalfunctiona(n)asfollows:a(n)=1,ifnisaperfectk-thpowernumber;0,otherwise.
InordertocompletetheproofofTheorem,weneedthefollowing:Lemma.
Foranyrealnumberx≥1,wehavetheasymptoticformulan≤x(n,pq)=1a(n)=x1k(p1)(q1)pq+Ox12k+.
Proof.
Letf(s)=∞n=1(n,pq)=1a(n)ns.
whereRe(s)>1.
FromtheEulerproductformula[4]andthemultiplicativepropertiesofa(n),wehavef(s)=P(P,pq)=11+a(Pk)Pks+a(P2k)P2ks+···=P1+1Pks+1P2ks11pks*11qks=ζ(ks)11pks*11qks,whereζ(s)istheRiemannzeta-function,andPdenotestheproductoverallprimes.
NowbyPerronformula[5]withs0=0,b=1k+1logx,T=x12k,H(x)=xandB(σ)=1σ1k,wehaven≤x(n,pq)=1a(n)=12πib+iTbiTζ(ks)(pks1)(qks1)(pq)ksxssds+O(x12k+ε).
Anarithmeticalfunctionandtheperfectk-thpowernumbers93Toestimatethemainterm,wemovetheintegrallinefromb=1k+1logxtoa=12k+1logxTherefore,12πib+iTbiT+a+iTb+iT+aiTa+iT+biTaiTf(s)xssds=Resf(s)xss,1k.
Notethatlims→1ζ(s)(s1)=1,wemayimmediatelygetResf(s)xss,1k=x1k11p11q.
Nowfromtheestimate12πia+iTb+iT+aiTa+iT+biTaiTf(s)xssdsx12k+,wecaneasilygetn≤x(n,pq)=1a(n)=x1k(p1)(q1)pq+Ox12k+.
Thisprovesthelemma.
Nowweprovethetheorem.
Fromthepropertiesofgeometricalseriesandthedenitionofepq(n),combiningthelemmawehaven≤xn∈Aepq(n)=α≤logpqxk|ααn≤x(pq)α(n,pq)=1a(n)=α≤logpqxkkαx(pq)kα1k(p1)(q1)pq+Ox(pq)kα12k+=kx1k(p1)(q1)pq∞n=1n(pq)nα>logpqxkα(pq)α+Ox12k+=kx1k(p1)(q1)pq∞n=1n(pq)n1(pq)logpqxk∞α=1α+logpqxk(pq)α+Ox12k+94RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII=kx1k(p1)(q1)pqap,q+O(x1klogx)+Ox12k+=(p1)(q1)pqap,qkx1k+x12k+,whereap,q=∞n=1n(pq)nisacomputablepositiveconstant.
ThiscompletestheproofofTheorem.
References[1]F.
Smarandache.
Onlyproblems,notSolutions.
XiquanPubl.
House,Chicago,1993.
[2]LvChuan.
Anumbertheoreticfunctionanditsmeanvalue,ResearchonSmarandacheproblemsinnumbertheory(collectedpapers).
Hexis,2004,33-35.
[3]ZhangWenpeng.
Anarithmeticfunctionandtheprimitivenumberofpowerp,ResearchonSmarandacheproblemsinnumbertheory(collectedpa-pers).
Hexis,2004,1-4.
[4]T.
M.
Apostol.
IntroductiontoAnalyticNumberTheory.
Springer-Verlag,1976.
[5]PanChengdongandPanChengbiao.
ElemensoftheanalyticnumberTheory.
Beijing,SciencePress.
1991.
ANARITHMETICALFUNCTIONANDITSMEANVALUEFORMULARenZhibinandZhaoXiaopengDepartmentofMathematics,WeinanTeacher'sCollege,Weinan,Shaanxi,P.
R.
ChinaAbstractInthispaper,weusetheelementarymethodstostudythemeanvaluepropertiesofanarithmeticalfunction,andgiveaninterestingasymptoticformulaforit.
Keywords:Arithmeticalfunction;Meanvalue;Asymptoticformula.
§1.
IntroductionLetkbeaxedpositiveinteger.
Foranypositiveintegern,wedenethearithmeticalfunctionbk(n)asfollows:bk(n)=maxm|mi=1ik≤n,n∈N.
Thatis,bk(n)isthegreatestpositiveintegermsuchthatmi=1ik≤n.
Forexample,b2(1)=1,b2(2)=1,b2(3)=1,b2(4)=1,b2(5)=2,Infact,fromthedenitionofbk(n)weknowthatif1≤n1,wehavetheasymptoticformulan≤xbk(n)=(k+1)k+2k+1k+2xk+2k+1+O(x).
FromtheTheorem,wemayimmediatelydeducethefollowingtwoCorollaries96RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIICorollary1.
Foranyrealnumberx>1,wehavetheasymptoticformulan≤xb1(n)=2323x32+O(x).
Corollary2.
Foranyrealnumberx>1,wehavetheasymptoticformulan≤xb2(n)=3434x43+O(x).
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofthetheorem.
Firstwedenefk(n)=ni=1ik.
Foranyrealnumberx>1,itisclearthatthereexistsoneandonlyonepositiveintegerNsuchthatfk(N)≤x0isaconstant.
100RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII§2.
TwosimplelemmasTocompletetheproofofthetheorem,weneedthefollowingtwosimplelemmas:Lemma1.
Letn!
=pα11pα22···pαkkdenotesthefactorizationofnintoprimepowers.
Thenwehavethecalculateformulab(n!
)=b(pα11)·b(pα22b(pαkk)=pord(p1)1·pord(p2)2pord(pk)2,wheretheordfunctionisdenedas:ord(pi)=1,ifαiisodd,0,ifαiiseven.
Proof.
Seereference[2].
Lemma2.
Foranyrealnumberx≥2,wehavetheasymptoticformulaθ(x)=p≤xlnp=x+OxexpAln35x(lnlnx)15.
Proof.
Seereferences[3]or[4].
§3.
ProofofthetheoremInthissection,weshallcompletetheproofofTheorem.
FirstfromLemma1wehavelnb(n!
)=lnb(pα11)·b(pα22b(pαkk)=p≤n2ord(p)lnp=n2√n,thenwehavetheidentityS(n)=P(n).
whereP(n)denotesthegreatestprimedivisorofn.
Proof.
Fromtheprimepowersfactorizationofn,wemayimmediatelygetpα11pα22···pαrr√n}.
UsingtheEuler'ssummationformula(seereference[2]),wemaygetn∈AS(n)n≤x√nlnn=x1√tlntdt+x1(t[t])(√tlnt)dt+√xlnx(x[x])x32lnx.
Similarly,fromtheAbel'ssummationforumulawealsohaven∈BS(n)=n≤xP(n)>√nP(n)=n≤√xn≤p≤xnp=n≤√x√x≤p≤xnp+On≤√xn≤p≤xn√x=n≤√xxnπxn√xπ(√x)xn√xπ(s)ds+Ox32lnx,whereπ(x)denotesallthenumbersofprimeswhichisnotexceedingx.
Notethatπ(x)=xlnx+Oxln2x.
OntheSmarandachefunction105Usingtheaboveasymptoticformula,wehave√x≤p≤xnp=xnπxn√xπ(√x)xn√xπ(s)ds=12x2n2lnx/n12xln√x+Ox2n2ln2x/n+Oxln2√x+Ox2n2ln2x/nxln2√x.
Consideringthefollowingn≤√x1n2=ζ(2)+O(1x).
Hencewehaven≤√xx2n2lnxn=n≤ln2xx2n2lnxn+Oln2x≤n≤√xx2n2lnxn=π26x2lnx+Ox2ln2xandn≤√xx2n2ln2xn=Ox2ln2x.
Combiningalltheabove,wemayimmediatelydeducethatn≤xS(n)=n∈AS(n)+n∈BS(n)=π2x212lnx+Ox2ln2x.
ThiscompletestheproofofLemma2.
§3.
ProofofthetheoremInthissection,weshallcompletetheproofofTheorem.
FirstapplyingtheAbel'ssummation,andnotethattheresultsofLemma1andLemma2,wemayhaven≤xS(n)n=1xn≤xS(n)+x11t2n≤tS(t)dt106RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII=1xπ2x212lnx+Ox2ln2x+x11t2π2t212lnt+Ot2ln2tdt=π212xlnx+Oxln2x+x1π212lntdt+Ox11ln2tdt=π26xlnx+Oxln2x.
ThiscompletestheproofofTheorem.
References[1]LiHailongandZhaoXiaopeng.
OntheSmarandachefunctionandthek-throotsofapositiveinteger.
Hexis,119-122,2004.
[2]TomMA.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
MEANVALUEOFTHEK-POWERCOMPLEMENTSEQUENCESFengZhiyuFengxiangTeacher'sSchool,Fengxiang,Shaanxi,P.
R.
ChinaAbstractForanyprimep≥3andanyxedintegerk≥2,letep(n)denotethelargestexponentofpowerpwhichdividesn,a(n,k)denotesthek-powercomplementnumberofn.
Inthispaper,westudythepropertiesofthesequenceep(a(n,k)),andgiveaninterestingasymptoticformulaforit.
Keywords:Largestexponent;Meanvalue;Asymptoticformula.
§1.
IntroductionLetp≥3beaprime,ep(n)denotethelargestexponentofpowerpwhichdividesn.
Foranyxedintegerk≥2,leta(n,k)denotek-powercomple-mentsequenceofn.
Thatis,a(n,k)isthesmallestpositiveintegersuchthatna(n,k)isaperfectk-power.
Inproblem29ofreference[1],professorF.
Smarandachaskedustostudythepropertiesofthissequences.
Aboutthisproblem,somepeoplehadstudieditbefore,andmadesomeprogress,seeref-erence[2].
Themainpurposeofthispaperisusingtheanalyticmethodtostudythepropertiesofthesequenceep(a(n,k)),andgiveaninterestingasymptoticformulaforitsmeanvalue.
Thatis,weshallprovethefollowing:Theorem.
Letp≥3beaprimeandk≥2axedpositiveinteger.
Thenforanyrealnumberx≥1,wehavetheasymptoticformulan≤xep(a(n,k))=(k1)pkkpk1+1(pk1)(p1)x+Ox1/2+ε,whereεdenotesanyxedpositivenumber.
FromthisTheoremwemayimmediatelydeducethefollowingtwocorol-laries:Corollary1.
Letp≥3beaprime.
Thenforanyrealnumberx≥1,wehavetheasymptoticformulan≤xep(a(n,2))=1p1x+Ox1/2+ε.
108RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIICorollary2.
Letp≥3beaprime.
Thenforanyrealnumberx≥1,wehaven≤xep(a(n,3))=2p+1p2+p+1x+Ox1/2+ε.
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofTheorem.
Infact,foranycomplexnumberswithRe(s)>1,wedenetheDirichletseriesf(s)=∞n=1ep(a(n,k))ns.
Fromthedenitionsofep(n)anda(n,k),andapplyingtheEulerproductformula(Seereference[4])wehavef(s)=∞α=0k1β=1kβp(αk+β)s∞n1=1(n1,p)=11(n1)s=k∞α=0k1β=11p(αk+β)s∞n1=1(n1,p)=11(n1)s∞α=01pαksk1β=1βpβs∞n1=1(n1,p)=11(n1)s=k11pks1ps1pks11psζ(s)11ps111pks1ps1pks11psk1pksζ(s)=p(k1)s1[(k1)psk](pks1)(ps1)ζ(s)+k1pks1ζ(s)=(k1)pkskp(k1)s+1(pks1)(ps1)ζ(s),whereζ(s)istheRiemannzeta-function.
Obviously,wehaveep(a(n,k))≤klogpn≤klnn∞n=1ep(a(n,k))nσ≤kσ1,whereσistherealpartofs.
ThereforebyParron'sformula(Seereference[3])wecangetn≤xep(a(n,k))ns0=12πib+iTbiTf(s+s0)xssds+OxbB(b+σ0)T+Ox1σ0H(2x)min1,lnxT+Oxσ0H(N)min1,xxMeanvalueofthek-powercomplementsequences109whereNisthenearestintegertox,x=|xN|.
Takings0=0,b=32,H(x)=klnx,B(σ)=kσ1,wehaven≤xep(a(n,k))=12πi32+iT32iTζ(s)R(s)xssds+O(x32+εT),whereR(s)=(k1)pkskp(k1)s+1(pks1)(ps1).
Toestimatethemainterm12πi32+iT32iTζ(s)R(s)xssds,wemovetheintegrallinefroms=32±iTtos=12±iT.
Thistime,thefunctiong(s)=ζ(s)R(s)xsshasasimplepolepointats=1,andtheresidueis(k1)pkkpk1+1(pk1)(p1)x.
Sowehave12πi32+iT32iT+12+iT32+iT+12iT12+iT+32iT12iTζ(s)R(s)xssds=(k1)pkkpk1+1x(pk1)(p1).
Notethattheestimate12πi12+iT32+iT+32iT12iT+12iT12+iTζ(s)(k1)pkskp(k1)s+1(pks1)(ps1)xssdsx32+εT.
TakingT=x,fromtheaboveformulawemayimmediatelygettheasymp-toticformulan≤xep(a(n,k))=(k1)pkkpk1+1x(pk1)(p1)+Ox1/2+ε.
ThiscompletestheproofofTheorem.
110RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIReferences[1]F.
Smarandache.
OnlyProblems,Notsolutions.
Chicago:XiquanPub-lishingHouse,1993.
[2]ZhangWenpeng.
ReseachonSmarandacheproblemsinnumbertheory.
HexisPublishingHouse,2004.
[3]PanChengdongandPanChengbiao.
FoundationofAnalyticNumberTheory.
Beijing:Sciencepress,1997.
[4]ApostolTM.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
ONTHEM-THPOWERFREENUMBERSEQUENCESChenGuohuiDepartmentofMathematics,HainanNormalUniversity,Haikou,Hainan,P.
R.
ChinaAbstractInthispaper,weusingtheelementarymethodstostudythearithmeticalproper-tiesofthem-thpowerfreenumbersequences,andgivesomeinterestingidenti-tiesforit.
Keywords:m-freenumbersequence;Dirichlet'sseries;Identity.
§1.
IntroductionForanypositiveintegernandmwithm≥2,wedenethem-thpowerfreenumberfunctionam(n)ofnasfollows:Ifn=pα11pα22···pαssistheprimepowersdecompositionofn,thenthem-thpowerfreenumberfunctionofnisthefunction:am(n)=pβ11pβ22···pβss,whereβi=αi,ifαi≤m1,andβi=0,ifαi≥m.
Foranypositiveintegerk,wealsodenethearithmeticalfunctionδk(n)asfollows:δk(n)=max{d∈N|d|n,(d,k)=1},ifn=0,0,ifn=0.
LetAdenotesthesetofallthepositiveintegersnsuchthattheequationam(n)=δk(n).
Thatis,A={n∈N,am(n)=δk(n)}.
Inthispaper,weusingtheelementarymethodstostudytheconvergentpropertiesoftheDirichletseriesinvolvingthesetA,andgiveaninterestingidentity.
Thatis,weshallprovethefollowingconclusion:Theorem.
Letm≥2beapositiveinteger.
Thenforanyrealnumbers>1,wehavetheidentity:∞n=1n∈A1ns=ζ(s)ζ(ms)p|kpmsp(m1)s+1pms1,whereζ(s)istheRiemannzeta-function.
Fromthistheoremwemayimmediatelydeducethefollowing:112RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIICorollary.
LetB={n∈N,a2(n)=δk(n)}andC={n∈N,a3(n)=δk(n)},thenwehavetheidentities:∞n=1n∈B1n2=15π2p|kp4p2+1p41and∞n=1n∈C1n2=30521π4p|kp6p4+1p61.
§2.
ProofofthetheoremInthissection,wewillcompletetheproofofthetheorem.
First,wedenethearithmeticalfunctiona(n)asfollows:a(n)=1,ifn∈A,0,ifotherwise.
Foranyrealnumbers>0,itisclearthat∞n=1n∈A1ns=∞n=1a(n)ns1,thus∞n=1n∈A1nsisalsoconvergentifs>1.
Nowletn=pα11pα22···pαssdenotesthefactorizationnintoprimepowers.
Thenfromthedenitionofam(n)andδk(n),weknowthatifαj≥mforsomej,thenam(pαjj)=1.
Soonlywhenαi≤m1and(pi,k)=1foralli,theequationam(n)=δk(n)hassolution.
If(pj,k)=1forsomej,thentheequationam(pαjj)=δk(pαjj)hassolutionifandonlyifαj≥m.
SofromtheEulerproductformula(see[1]),wehave∞n=1n∈A1ns=p1+a(p)ps+a(p2)p2sa(pm1)p(m1)s+···=pk1+a(p)ps+a(p2)p2sa(pm1)p(m1)s*p|k1+a(p)pms+a(p2)p(m+1)s+a(p3)p(m+2)s+···=pk1+1ps+1p2s1p(m1)sOnthem-thpowerfreenumbersequences113*p|k1+1pms+1p(m+1)s+1p(m+2)s+···=ζ(s)ζ(ms)p|kpmsp(m1)s+1pms1,whereζ(s)istheRiemannzeta-functionandpdenotestheproductoverallprimes.
ThiscompletestheproofofTheorem.
References[1]TomM.
Apostol.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
[2]F.
Smarandache.
Onlyproblems,notsolutions,XiquanPublishingHouse,Chicago,1993.
ANUMBERTHEORETICFUNCTIONANDITSMEANVALUEDuJianliSchoolofScience,Xi'anShiyouUniversity,Xi'an,Shaanxi,P.
R.
ChinaAbstractLetpandqaretwodistinctprimes,epq(n)denotesthelargestexponentofpowerpqwhichdividesn,andletbpq(n)=t|nepqntepq(t).
Inthispaper,weusingtheanalyticmethodstostudythemeanvlauepropertiesofthebpq(n),andgiveaninterestingasymptoticformulaforit.
Keywords:Largestexponent;Meanvalue;Asymptoticformula.
§1.
IntroductionLetpandqaretwodistinctprimes,epq(n)denotesthelargestexponentofpowerpqwhichdividesn.
Inproblem68of[3],professorF.
Smarandacheaskedustostudythepropertiesofthesequenceep(n).
Aboutthisprob-lem,somepeoplehadstudiedit,andobtainedaseriesinterestingresults,seereferences[2].
Inthispaper,wedenearithmeticalfunctionbpq(n)=t|nepqntepq(t),thenweusingtheanalyticmethodstostudythemeanvaluepropertiesofbpq(n),andgivegiveasharpasymptoticformulaforit.
Thatis,weshallprovethefollowing:Theorem.
Letpandqaretwodistinctprimes,thenforanyrealnumberx≥1,wehavetheasymptoticformulan≤xbpq(n)=xlnx(pq1)2+12γpq+2pqγ2pqln(pq)(pq1)3x+Ox12+ε,whereεisanyxedpositivenumber,andγistheEulerconstant.
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofTheorem.
Foranycomplexs,wedenetheDirichlet'sseriesf(s)=∞n=1epq(n)nsandg(s)=∞n=1bpq(n)ns.
116RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIItisclearthatg(s)=f2(s).
RenGanglianhasgiven(inapapertoappear)f(s)=ζ(s)(pq)s1,whereζ(s)istheRiemannzeta-function.
Sowecanobtainthatg(s)=ζ2(s)((pq)s1)2.
Obviously,wehavebpq(n)≤logpqn≤nlnn∞n=1bpq(n)nσ≤1σ2,whereσ(≥2)istherealpartofs.
Therefore,byParron'sformula(Seerefer-ence[4])wecangetn≤xbpq(n)ns0=12πib+iTbiTζ2(s+s0)R(s+s0)xssds+OxbB(b+σ0)T+Ox1σ0H(2x)min1,logxT+Oxσ0H(N)min1,xx,whereNbethenearestintegertox,x=|xN|.
LetR(s)=1((pq)s1)2.
Takings0=0,b=52,H(x)=xlnx,B(σ)=1σ2,wehaven≤xbpq(n)=12πi52+iT52iTζ2(s)R(s)xssds+O(x32+εT).
Toestimatethemainterm12πi52+iT52iTζ2(s)R(s)xssds,wemovetheintegrallinefroms=52±iTtos=12±iT.
Thistime,thefunctionζ2(s)R(s)xsshasasecondorderpolepointats=1,andtheresidueisxlnx(pq1)2+12γpq+2pqγ2pqln(pq)(pq1)3x,whereγisEulerconstant.
Sowehave12πi52+iT52iT+12+iT52+iT+12iT12+iT+52iT12iTζ2(s)R(s)xssds=xlnx(pq1)2+12γpq+2pqγ2pqln(pq)(pq1)3x.
Anumbertheoreticfunctionanditsmeanvalue117TakingT=x2,wecaneasilyobtain12πi12+iT52+iT+52iT12iTζ2(s)R(s)xssdsx12+εand12πi12iT12+iTζ2(s)R(s)xssdsx12+ε.
Sofromtheaboveformula,wemayimmediatelygettheasymptoticformulan≤xbpq(n)=xlnx(pq1)2+12γpq+2pqγ2pqln(pq)(pq1)3x+Ox1/2+ε.
ThiscompletestheproofofTheorem.
References[1]ApostolTM.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
[2]ZhangWenpeng.
ReseachonSmarandacheproblemsinnumbertheory.
HexisPublishingHouse,2004.
[3]F.
Smarandache.
OnlyProblems,Notsolutions.
Chicago:XiquanPub-lishingHouse,1993.
[4]PanChengdongandPanChengbiao.
FoundationofAnalyticNumberTheory.
Beijing:Sciencepress,1997.
ONTHEADDITIVEK-THPOWERPARTRESIDUEFUNCTIONZhangShengshengShaanxiTechnicalCollegeofFinanceandEconomics,Xianyang,Shaanxi,P.
R.
ChinaAbstractSimilartotheSmarandachek-thpowercomplements,wedenetheadditivek-thpowerpartresiduefk(n)ofnisthesmallestnonnegativeintegersuchthatnfk(n)isaperfectk-thpower.
Themainpurposeofthispaperisusingtheelementarymethodstostudythemeanvaluepropertiesoffk(n)andd(fk(n)),andgivetwointerestingasymptoticformulaeforthem.
Keywords:Additivek-thpowerpartresiduefunction;Meanvalue;Asymptoticformula.
§1.
IntroductionandresultsForanypositiveintegern,theSmarandachek-thpowercomplementsbk(n)isthesmallestpositiveintegerbk(n)suchthatnbk(n)isacompletek-thpower(seeproblem29of[1]).
SimilartotheSmarandachek-thpowercomplements,XuZhefengin[2]denedtheadditivek-thpowercomplementsak(n)asfol-lows:ak(n)isthesmallestnonnegativeintegersuchthatn+ak(n)isacom-pletek-thpower.
Similarly,wewilldenetheadditivek-thpowerpartresiduefk(n)asfol-lowing:foranypositiveintegern,fk(n)=min{r|0≤r=nmk,m∈N}.
Forexample,ifk=2,wehavetheadditivesquarepartresiduesequences{f2(n)}(n=1,2,asfollowing:0,1,2,0,1,2,3,4,0,1,2,3,4,5,6,0,1,2,Inthispaper,weusetheelementarymethodstostudythemeanvalueproper-tiesoffk(n)andd(fk(n))(whered(n)istheDirichletdivisorfunction),andobtainsomeinterestingasymptoticformulaeforthem.
Thatis,wewillprovethefollowingconclusions:Theorem1.
Foranyrealnumberx≥3andintegerk≥2,wehavetheasymptoticformulan≤xfk(n)=k22(2k1)x21k+Ox22k.
120RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIITheorem2.
Foranyrealnumberx≥3andintegerk≥2,wealsohavetheasymptoticformulan≤xd(fk(n))=11kxlnx+2γ+lnk2+1kx+Ox11klnx,whereγistheEulerconstant.
§2.
SomeLemmasTocompletetheproofofthetheorems,weneedfollowingtwoLemmas.
FirstwehaveLemma1.
Foranyrealnumberx≥3,wehavetheasymptoticformulan≤xd(n)=xlnx+(2γ1)x+O(x12),whereγistheEulerconstant.
Proof.
Seereference[3].
Lemma2.
Foranyrealnumberx≥3andanynonnegativearithmeticalfunctionh(n)withh(0)=0,wehavetheasymptoticformulan≤xh(fk(n))=M1t=1n≤g(t)h(n)+Ox≤g(M)h(n),whereg(t)=k1i=1kitiandM=[x1k],[x]denotesthegreatestintegernotexceedingx.
Proof.
Foranyrealnumberx≥1,letMbeaxedpositiveintegersuchthatMk≤x1,itisclearthatf(s)isabsolutelyconvergent.
SofromthelemmaandtheEuler'sproductformula[2]wehavef(s)=p1+D(Ak(p))ps+D(Ak(p2))p2s+···=p1+D(pk1)ps+D(pk2)p2sD(1)pks+···=p1+2v(pk1)ps+2v(pk1)p2s2v(1)pks+···Anarithmeticalfunctionandthek-thpowercomplements125=p1+2ps+2p2s1pks+2p(k+1)s+···=ζ(s)ζ(ks)p1+1ps2pks=ζ2(s)ζ(ks)ζ(2s)p12pks+p(k1)s,whereζ(s)istheRiemannzeta-function.
ThereforebyPerron'sformula[3]withs0=0,b=2,T=32,wehaven≤xD(Ak(n))=12πi2+iT2iTζ2(s)ζ(ks)ζ(2s)R(s)xssds+O(x12+ε),whereR(s)=p12pks+p(k1)s.
Toestimatethemainterm12πi2+iT2iTζ2(s)ζ(ks)ζ(2s)R(s)xssds,wemovetheintegrallinefroms=2±iTtos=12±iT,thenthefunctionζ2(s)ζ(ks)ζ(2s)R(s)xsshaveasecondorderpolepointats=1withresidueζ(k)ζ(2)xlnxp12pk+pk1+C(k)x,whereC(k)isacomputableconstant.
Sowehave12πi2+iT2iT+12+iT2+iT+2+iT12iT+2iT12iTζ(s)ζ(ks)R(s)xssds=6ζ(k)xlnxπ2p12pk+pk1+C(k)x.
Wecaneasilygettheestimate12πi12+iT2+iT+2+iT12iT+2iT12iTζ(s)ζ(ks)R(s)xssdsx12+ε.
Nownotethatζ(2)=π2/6,wemayimmediatelyobtaintheasymptoticfor-mulan≤xD(Ak(n))=6π2ζ(k)xlnxp12pk+pk1+C(k)x+O(x12+),126RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIIwhereC(k)isacomputableconstant.
Thiscompletestheproofofthetheorem.
References[1]F.
Smarandache.
OnlyProblems,Notsolutions.
Chicago:XiquanPub-lishingHouse,1993.
[2]ApostolTM.
IntroductiontoAnalyticNumberTheory.
NewYork:Springer-Verlag,1976.
[3]PanChengdongandPanChengbiao.
FoundationofAnalyticNumberTheory.
Beijing:Sciencepress,1997.
[4]ZhuWeiyi.
Onthek-powercomplementandk-powerfreenumberse-quence.
SmarandacheNotionsJournal,2004,14:66-69.
[5]LiuHongyanandLouYuanbing.
ANoteonthe29-thSmarandache'sproblem.
SmarandacheNotionsJournal,2004,14:156-158.
ONTHETRIANGLENUMBERPARTRESIDUEOFAPOSITIVEINTEGERJiYongqiangDanfengTeacher'sSchool,Shangluo,Shaanxi,P.
R.
ChinaAbstractForanypositiveintegern,leta(n)denotesthetrianglenumberpartresidueofn.
Thatis,a(n)=nk(k+1)2,wherekisthegreatestpositiveintegersuchthatk(k+1)2≤n.
Inthispaper,weusingtheelementarymethodstostudythemeanvaluepropertiesofthesequences{a(n)},andgivetwointerestingasymptoticformulaeforit.
Keywords:Thetrianglenumberpartresidue;Meanvalue;Asymptoticformula.
§1.
IntroductionandresultsForanypositiveintegern,leta(n)denotesthetrianglenumberpartresidueofn.
Thatis,a(n)=nk(k+1)2,wherekisthegreatestpositiveintegersuchthatk(k+1)2≤n.
Forexample,a(1)=0,a(2)=1,a(3)=0,a(4)=1,a(5)=2,a(6)=0,Inreference[1],American-RomaniannumbertheoristProfessorF.
Smarandacheaskedustostudythearithmeticalpropertiesofthissequences.
Aboutthisproblem,itseemsthatnonehadstudiedit,atleastwehavenotseenanyrelatedpapersbefore.
Inthispaper,weusingtheelementarymethodstostudythemeanvaluepropertiesofthissequences,andgivetwointerestingasymptoticformulaeforit.
Thatis,weshallprovethefollowing:Theorem1.
Foranyrealnumberx≥3,wehavetheasymptoticformulan≤xa(n)=√23x32+O(x).
Theorem2.
Foranyrealnumberx≥3,wealsohavetheasymptoticformulan≤xd(a(n))=12xlnx+2γ+ln232x+O(x23),whered(n)istheDirichletdivisorfunction(provided(0)=0),andγistheEulerconstant.
128RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYIINote:Itisclearthatifthereexistsameanvalueformulaforanyarithmeti-calfunctionf(n),thenusingourmethodswecanalsoobtainanasymptoticformulaforn≤xf(a(n)).
§2.
ProofofthetheoremsInthissection,weshallcompletetheproofofthetheorems.
FirstweproveTheorem1.
Foranyrealnumberx≥3,letMbeaxedpositiveintegersuchthatM(M+1)2≤x0isanyxedrealnumber,andqdenotestheproductoverallprimeq.
§2.
ProofofthetheoremInthissection,weshallcompletetheproofofTheorem.
Leta(n)denotesthecharacterfunctionofk-fullnumber.
Thatis,a(n)=1,ifnisak-fullnumber;0,otherwise.
Itisclearthatn≤xn∈A(n,p)=11=n≤x(n,p)=1a(n).
LetDirichletseriesf(s)=∞n=1(n,p)=1a(n)ns.
ItisclearthatthisseriesisconvergentifRes>1.
FromtheEulerproductformula[4]andthedenitionofa(n),wehavef(s)=q=p1+a(qk)qks+a(qk+1)q(k+1)s+···=q=p1+1qks·111qs=ζ(ks)ζ(2ks)11pksp(k1)s+1q1+1(qks+1)(qs1),(1)whereζ(s)istheRiemannzeta-function.
Anarithmeticalfunctionandthek-fullnumbersequences133Obviously,wehave|a(n)|≤1,∞n=1a(n))nσ≤1σ1k,whereσistherealpartofs.
Therefore,bythePerron'sformula[4]wehaven≤x(n,p)=1a(n)=12πib+iTbiTζ(ks)ζ(2s)R(s)xssds+O(x12k+ε),whereR(s)=11pksp(k1)s+1q1+1(qks+1)(qs1).
Nowmovingtheintegrallinefrombtoa=12k,wehave12πib+iTbiT+a+iTb+iT+aiTa+iT+biTaiTf(s)xssds=Resf(s)xss,1k.
(2)Notethatlims→1kζ(ks)s1k=1,wemayimmediatelygetResf(s)xss,1k=kx1kζ(2)R1k.
(3)Combining(1),(2),(3)andthefollowingestimates12πiaiTa+iTf(s)xssdsx12k+and12πia+iTb+iT+biTaiTf(s)xssdsx12k+wecaneasilygetn≤x(n,p)=1a(n)=C(p,k)x1k+Ox12k+,whereC(p,k)=6kπ211ppk1k+1q1+1(q+1)(q1k1).
Basedonthedenitionofep(n)andtheaboveestimate,wehaven≤xn∈Aep(n)=k≤α≤logpxαn≤xpα(n,p)=1a(n)134RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII=k≤α≤logpxαxpα1kC(p,k)+Oxpα12k+=C(p,k)x1kk≤α≤logpxαpαk+Ox12k+k≤α≤logpxαpαk=C(p,k)x1k∞n=1npnkα>logpxαpαkα1),thenf(s)=p1+am(p)ck(p)psam(pm)ck(pm)pms+am(pm+1)ck(pm+1)p(m+1)sam(p2m)ck(p2m)p2msam(pim+1)ck(pim+1)p(im+1)sam(pk1)ck(pk1)p(k1)s=p1+1psm+1p2sm(1+1ps1p(m2)s)+1p(m+1)s2m(1+1ps1p(m1)s)1p((i1)m+1)sim+k1j=im+1p(i+1)mpjs=ζ(sm)ζ(2(sm))p1+(p(m1)s1)+(pms1)i1j=1pjm+sp(jm+1)s(psm+1)(pmsp(m1)s)+k1j=im+1pim+spjspsm+1.
Obviously,wehavetheinequality|am(n)ck(n)|≤n2,∞n=1am(n)ck(n)nσm+1istherealpartofs.
SobyPerronformula[7]wehave:n≤xa(n)ns0=12πib+iTbiTf(s+s0)xssds+OxbB(b+σ0)T+Ox1σ0H(2x)min1,logxT+Oxσ0H(N)min1,xT||x||.
Onthehybridmeanvalueofsomespecialsequences2139whenNisthenearestintegertox,N=x12,x=|xN|.
Takings0=0,b=m+2,T=x32,H(x)=x2,B(σ)=1σm1,wehave:n≤xam(n)ck(n)=12iπm+2+iTm+2iTζ(sm)ζ(2(sm))R(s)xssds+O(xm+12+ε),whereR(s)=p1+p(k2)s1(psm+1)(p(k1)sp(k2)s),m≥k;p1+(p(m1)s1)+(pms1)i1j=1pjm+sp(jm+1)s(psm+1)(pmsp(m1)s)+k1j=im+1pim+spjspsm+1,m1),thenf1(s)=p1+pmpm1pspmpm1pms+p2mp2m1p(m+1)sp2mp2m1p2msp(i+1)mp(i+1)m1p(im+1)sp(i+1)mp(i+1)m1p(k1)s=p1+1psm+1p2sm1+1ps1p(m2)s1psm+11+1ps1p(m1)s1p(m+1)s2m1+1ps1p(m1)s1p(m+1)s2m+11+1ps1p(m1)s1p((i1)m+1)sim1+1ps1p(m1)s1p((i1)m+1)sim+11+1ps1p(m1)s+k1j=im+1p(i+1)mp(i+1)m1pjs=ζ(sm)ζ(2(sm))p1+(p(m1)s1)+(pms1)H(s)(psm+1)(pmsp(m1)s)+G(s)psm+1,whereH(s)=i1j=1(pjm+s+1pjm+sp(jm+1)s1p)andG(s)=k1j=im+1pim+spim+s1pjs.
ByPerronformula[7]andthemethodofprovingTheorem1,wealsohaven≤xφ(am(n)ck(n))=12iπm+2+iTm+2iTζ(sm)ζ(2(sm))R(s)xssds+O(xm+12+ε),whereR(s)=p1+p(p(k2)s1)(p(k1)s1)p(psm+1)(p(k1)sp(k2)s),m≥k;p1+(p(m1)s1)+(pms1)H(s)(psm+1)(pmsp(m1)s)+G(s)psm+1,m1).
Thenforanyxedpositiveintegermandanyrealnumberx≥1,wehavetheasymptoticformulan≤xn∈Aδm(f(n))=R(1)ζ(k)x+Ox12+ε,whereζ(k)istheRiemann-zetafunction,εdenotesanyxedpositivenumber,andR(s)=p|qpm11pks11pspβqpmβ1istherealpartofs.
SobyPerronformulawehaven≤xa(n)ns0=12iπb+iTbiTf(s+s0)xssds+OxbB(b+σ0)T+Ox1σ0H(2x)min(1,logxT)+Oxσ0H(N)min(1,x||x||),whereNisthenearestintegertox,x=|xN|.
Takings0=0,b=2,T=x32,H(x)=x2,B(σ)=1σ1,wehaven≤xn∈Aδm(f(n))=12iπ2+iT2iTζ(s)ζ(ks)R(s)xssds+O(x12+ε),whereR(s)=p|qpm11pks11pspβqpmβ≥k1+pps+p2p2spk1p(k1)s*pβqpmβToestimatethemainterm12iπ2+iT2iTζ(s)ζ(ks)R(s)xssds,Onthemeanvalueofanewarithmeticalfunction147wemovetheintegrallinefroms=2±iTtos=12±iT.
Thistime,thefunctionf(s)=ζ(s)xsζ(ks)sR(s)hasasimplepolepointats=1withresidueR(1)ζ(k)x.
Sowehave12iπ2+iT2iT+12+iT2+iT+12iT12+iT+2iT12iTζ(s)xsζ(ks)sds=R(1)xζ(k).
Wecaneasilygettheestimate12πi12+iT2+iT+2iT12iTζ(s)xsζ(ks)sds212ζ(σ1+iT)ζ(k(σ1+iT))x2Tdσx2T=x12and12πi12iT12+iTζ(s)xsζ(ks)sdsT0ζ(1/2+it)ζ(k/2+kit)x12t+1dtx12+ε.
Combiningtheaboveestimateswehaven≤xn∈Aδm(f(n))=R(1)ζ(k)x+Ox12+ε.
ThiscompletestheproofofTheorem.
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Thisbookcontains34papers,mostofwhichwerewrittenbyparticipantstotheFirstNorthwestNumberTheoryConfe-renceheldinShangluoTeacher'sCollege,China,inMarch,2005.
InthisConference,severalprofessorsgaveatalkonSmarandacheProblemsandmanyparticipantslecturedonthembothextensivelyandintensively.
Allthesepapersareoriginalandhavebeenrefereed.
ThethemesofthesepapersrangefromthemeanvalueorhybridmeanvalueofSmarandachetypefunctions,themeanvalueofsomefamousnumbertheoreticfunctionsactingontheSmaran-dachesequences,totheconvergencepropertyofsomeinfiniteseriesinvolvingtheSmarandachetypesequences.
(TheEditors)ListoftheContributorsZhangWenpengYiYuanXuZhefengXueXifengRenGanglianDingLipingLiJieLiuYanniMaJinpingLiZhanhuWangTingYangHaiRenDongmeiLuYamingLiuDuansenLiJunzhuangLiChaoYangCundianZhaoJianWangNianliangGaoNanLouYuanbingYangQianliYangMingshunRenZhibinZhaoXiaopengWangyongxingFengZhiyuChenGuohuiDuJianliFuRuiqinZhangShengshengHuangWeiJiYongqiangZhaoJiantangLiYanshengMaJunqing