MATHEMATICS OF COMPUTATION Volume Number Pages S XX ODD PERFECT NUMBERS ARE GREATER THAN PASCAL OCHEM AND MICHA EL RAO Abstract - PDF document

MATHEMATICS OF COMPUTATION Volume  Number  Pages  S XX ODD PERFECT NUMBERS ARE GREATER THAN   PASCAL OCHEM AND MICHA EL RAO Abstract
MATHEMATICS OF COMPUTATION Volume  Number  Pages  S XX ODD PERFECT NUMBERS ARE GREATER THAN   PASCAL OCHEM AND MICHA EL RAO Abstract

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MATHEMATICSOFCOMPUTATIONVolume00,Number0,Pages000{000S0025-5718(XX)0000-0ODDPERFECTNUMBERSAREGREATERTHAN101500PASCALOCHEMANDMICHAELRAOAbstract.Brent,Cohen,andteRieleprovedin1991thatanoddperfectnumberNisgreaterthan10300.WemodifytheirmethodtoobtainN�101500.WealsoobtainthatNhasatleast101notnecessarilydistinctprimefactorsandthatitslargestcomponent(i.e.divisorpawithpprime)isgreaterthan1062.1.IntroductionAnaturalnumberNissaidperfectifitisequaltothesumofitspositivedivisors(excludingN).ItiswellknownthatanevennaturalnumberNisperfectifandonlyifN=2k1(2k1)foranintegerksuchthat2k1isaMersenneprime.Ontheotherhand,itisalong-standingopenquestionwhetheranoddperfectnumberexists.Inordertoinvestigatethisquestion,severalauthorsgavenecessaryconditionsfortheexistenceofanoddperfectnumberN.EulerprovedthatN=pq2foraprimep,withp=e=1(mod4)andgcd(p;q)=1.MorerecentresultsshowedthatNmustbegreaterthan10300[1],itmusthaveatleast75primefactors(countingmultiplicities)[4],anditmusthaveatleast9distinctprimefactors[5].MoreoverthelargestprimefactorofNmustbegreaterthan108[3],andNmusthaveacomponentgreaterthan1020[2](i.e.Nmusthaveadivisorpawithpprime,andpa�1020).Weimproveinthispapersomeoftheseresults.InSection3weshowthatNmustbegreaterthan101500.WeuseforthistheapproachofBrentetal.[1],withamethodtoby-passdeadlockssimilartothemethodusedbyHare[4].Withaslightmodi cationoftheapproach,weshowthatNmusthaveatleast101primefactorsinSection4,andthatNmusthaveacomponentgreaterthan1062inSection5.Theseresultsareoutcomesofsomeimprovementsintheusedtechniques,andoffactorisatione orts.WediscussthatinSection6.2.PreliminariesLetnbeanaturalnumber.Let(n)denotethesumofthepositivedivisorsofn,andlet1(n)=(n) nbetheabundancyofn.Clearly,nisperfectifandonlyif1(n)=2.We rstrecallsomeeasyresultsonthefunctionsand1.Ifpisprime,(pq)=pq+11 p1,and1(p1)=limq!+11(pq)=p p1.Ifgcd(a;b)=1,then(ab)=(a)(b)and1(ab)=1(a)1(b).EulerprovedthatifanoddperfectnumberNexists,thenitisoftheformN=pm2wherep=e=1(mod4)andgcd(p;m)=1.Theprimepissaidtobethespecialprime.c\rXXXXAmericanMathematicalSociety1 2PASCALOCHEMANDMICHAELRAOManyresultsonoddperfectnumbersareobtainedusingthefollowingargument.SupposethatNisanoddperfectnumber,andthatpisaprimefactorofN.IfpqkNforaq�0,then(pq)j2N.Thusifwehaveaprimefactorp0�2of(pq),wecanrecurseonthefactorp0.Wemakeallsuppositionsforqupwegetacontradiction(e.g.pqisgreaterthanthelimitwewanttoprove).Moreover,since(pa)j(p)ifa+1jb+1,wecanonlysupposethatpqkNforqsuchthatq+1isprime.Majorchangesbetweentheapproachestogetthetheoremsarethesuppositionwemakeonthehypotheticaloddperfectnumber,theorderofexplorationofprimefactors,andcontradictionsweuse.3.SizeofanoddperfectnumberTheorem1.Anoddperfectnumberisgreaterthan101500.Weusefactorchainsasdescribedin[1]toforbidthefactorsinS=f127;19;7;11;331;31;97;61;13;398581;1093;3;5;307;17g,inthisorder.Thesechainsarecon-structedusingbranchings.Tobranchonaprimepmeansthatwesequentiallybranchonallpossiblecomponentspa.TobranchonacomponentpaforpprimemeansthatwesupposepakN,andthuspa(pa)j2Nsincegcd(pa;(pa))=1.Then,ifwedonotreachacontradictionatthispoint,werecursivelybranchonaprimefactorofNthathasnotyetbeenbranchedon.IfthereisnoknownotherfactorofN,wehaveasituationcalledroadblock,thatisdiscussedbelow.Twotypesofthelatterbranchingarealsodiscussedbelow.Inthissection,webranchontheoveralllargestavailableprimefactorandusethefollowingcontradictions:-Theabundancyofthecurrentnumberisstrictlygreaterthan2.-Thecurrentnumberisgreaterthan101500.Whenbranchingonaprimep,wehavetoconsidervariouscasesdependingonthemultiplicityofpinN.Westopwhenthemultiplicityaofpissuchthatpa�101500and,exceptinthecasesdescribedbelow,weconsideronlythemulti-plicitiesasuchthata+1isprime.Thisisbecause(pa)jp(a+1)1,soanycontradictionobtainedthankstothefactorsof(pa)whensupposingpakNalsogivesacontradictioninthecasep(a+1)1kN.Sopaisarepresentativeforallp(a+1)1,andtocomputelowerboundsontheabundancyorthesizetotestforcontradictions,wesupposethatthemultiplicityofpisexactlya.By-passingroadblocks.Aroadblockisasituationsuchthatthereisnocon-tradictionandnopossibilitytobranchonaprime.Thishappenswhenwehavealreadymadesuppositionsforthemultiplicityofalltheknownprimesandtheothernumbersarecomposites.WeuseamethodtocircumventroadblockssimilartotheoneusedbyHare[4].Thismethodrequirestoknowanupperboundontheabundancyofthecurrentnumberthatisstrictlysmallerthan2.Anobviousupperboundonthecontributionofthecomponentpatotheabundancyis1(p1)=p p1,butitmightnotalwaysensurethattheboundontheabundancyofthecurrentnumberisstrictlysmallerthan2.Inordertoobtaingoodenoughupperboundsontheabundancy,wedistinguishbetweenexactbranchingsandstandardbranchings.Exactbranchingsconcernthespecialcomponentp1,aswellas32,34,and72.Standardbranchingsconcerneverythingelse. ODDPERFECTNUMBERSAREGREATERTHAN1015003Inthecaseofanexactbranchingonpa,wesupposethatpakN,weuse1(pa)fortheabundancy,andweuseanadditionalcontradiction,occurringwhenpap-pearsatleasta+1timesinthefactorsofk=1(pq),where(pq11;:::;pqkk)isthesequenceofconsideredbranchings.Inthecaseofastandardbranchingonpa,wesupposethatp(a+1)1kNforat1,andweuse1(p1)=p p1asanupperboundontheabundancy.Duetotheseexactbranchings,wehavetoaddstandardbranchingson38,314,324,and78inordertocoverallpossibleexponentsfor3and7.Letusdetailthisforthebase3:wemakeexactbranchingson32and34,andstandardbranchingson38,314,324,and3p1foreveryprimep7.Thenthecase3m1kNformoddishandledby32ifm=3,by34ifm=5,by38if32jm,by314if35jm,by324if52jm,andby3p1ifpjm.Notethatwesupposethatthebranchingforthespecialprimep1isalwaysanexactbranching,sinceifp4k+1kNwithk1,thenthiscasewillbehandledbythestandardbranchingpq1,whereqisafactorof2k+1.Finallywehavetoconsiderabundancyofnonfactoredcomposites.WecheckthatthecompositeChasnofactorslessthan (weused =108forourcompu-tations),thusChasatmostjln(C) ln()kdi erentprimefactor,eachgreaterthan .ThustheabundancycontributedbyCisatmost 1bln(C) ln( )c.GivenaroadblockM,wecomputeanupperboundaontheabundancy.Ourmethodtoby-passtheroadblockonlyworksifa2.Thatiswhytheexactbranchingsweresuitablychosentoensurethata2foreveryroadblock.Supposethata2andthatthereisanoddperfectnumberNdivisiblebyM.LetpbethesmallestprimewhichdividesNandnotM.ThusNhasatleastta(p):=ln(2 a) ln( 1)distinctprimefactorswhichdonotdivideM.Eachofthesefactorshasmultiplicityatleast2,exceptforatmostone(special)primewithmultiplicityatleastone.Thusifp2a(p)1isgreaterthan101500 M,Nisclearlygreaterthan101500.Letb=maxnp:p2a(p)1101500 Mo,whichisde nedsincepp2a(p)1isstrictlygrowing.ToprovethatthereisnooddperfectnumberN101500suchthatMdividesN,webranchoneveryprimefactoruptobtorulethemout.WestarttobranchontheprimesinS,sincewealreadyhavegoodfactorchainsforthesesnumbers.WedonotbranchonaprimethatdividesMorthatisalreadyforbidden.Whenapplyingthismethod,wemightencounterotherroadblocks,becauseofcompositenumberorbecauseevery\produced"primealreadydividesM.Sowehavetoapplythemethodrecursively.Example.Anexampleofby-passtwonestedroadblocksisshowninFigure3.We rsttrytoruleout127asafactorandencounterasa rstroadblock127192,whichisacompositenumberwithnoknownfactorsandnofactorslessthan108.Here,M=127192127192&#x-0.8;͸䀀710807.Thiscompositenumberhasatmostln((127192)) ln(108)=50factorswhocontributetotheabundancyuptoatmostC=(1+108)501+6107.Asanupperboundontheabundancy,wethushavea=1(1271)1+61071:008.Wetryeverynumberuntilwe 4PASCALOCHEMANDMICHAELRAO127192=127192Roadblock192=312732=13131=2772=319Roadblock2Figure1.Exampleoftwonestedroadblockcircumvents.getta(220)=151and220301�10705�101500 M.So,togetaroundthisroadblock,wehavetobranchoneveryprimep220except127.Westartby19,whichisthenextnumberinS,andthenwegetstuckwithanotherroadblock(\Roadblock2").Here,M0=3272131192127192127192&#x-0.8;أ倀10814.Asanupperboundontheabundancy,wehavea0=132721311911271C.Wethushaveanupperbounda01:92522.Wetryeverynumberuntilwegetta0(2625)=101and2625201&#x-0.8;أ倀10687&#x-0.8;أ倀101500 M0.So,togetaroundthisroadblock,wehavetobranchoneveryprimepsuchthatp2625,except3,7,13,19and127.WecontinuetobranchonotherprimesinS,andthenoneveryotherprimessmallerthan2625.Thislastexampleshowsthatexactbranchingson32and72arenecessarysince131711311911271&#x-0.8;أ倀2.Noticealsotheexactbranchingonthespecialprime13.WhenNhasnofactorsinS.Finally,wehavetoshowthatifNhasnodivisorinS,thenN&#x-0.8;أ倀101500.Weusethefollowingargument,whichisanimprovedversionoftheargumentin[1].Foraprimepandanintegera,wede netheeciencyf(p;a)ofthecomponentpaasf(p;a)=ln(1(pa)) ln(pa).Theeciencyistheratiobetweenthecontributioninabundancyandthecontributioninsizeofthecomponentpa.Bothcontributionsaremultiplicativeincreasings,whichexplainsthelogarithms.Remark.ab=f(p;a)&#x-279;&#x.055;f(p;b).pq=f(p;a)&#x-278;&#x.153;f(q;a).Noticethatthebestwaytoreachabundancy2andtokeepNsmallistotakecomponentswithhighesteciencyf:Foreachallowedprimep,we ndthesmallestexponentasuchthat(pa)isnotdivisibleby4norafactorinS.Example:Considerp=23.(231),(232),(233),arerespectivelydivisibleby4,7,4.Sotheexponentof23isatleast4.Wesortthesecomponentspabydecreasingeciencyftogetanorderingp1;p2;p3;:::suchthatf(p1;a1)f(p2;a2)f(p3;a3):::.Theproduct200=1p p1=1:99785:::issmallerthan2,whereastheproduct200=1paisgreaterthan101735.4.TotalnumberofprimefactorsofanoddperfectnumberHareprovedthatanoddperfectnumberhasatleast75primefactors(countingmultiplicities)[4]. ODDPERFECTNUMBERSAREGREATERTHAN1015005Theorem2.Thetotalnumberofprimefactorsofanoddperfectnumberisatleast101.Weusethefollowingcontradictions:-Theabundancyofthecurrentnumberisstrictlygreaterthan2.-Thecurrentnumberhasatleast101primefactors.WeforbidthefactorsinS0=f3;5;7;11g,inthisorder.Webranchonthesmallestavailableprime.Westilluseacombinationofexactbranchings(forp1,32,and34)andstandardbranchings,asintheprevioussection.By-passingroadblocks.GivenaroadblockMwithatleastgnotnecessarilydis-tinctprimefactors,wecomputeanupperboundaontheabundancy,asdescribedintheprevioussection.Supposethata2andthatthereisanoddperfectnumberNdivisiblebyM.LetpbethesmallestprimewhichdividesNandnotM.ThusNhasatleastta(p)distinctprimefactorswhichdonotdivideM.Eachofthesefactorshasmultiplicityatleast2,exceptforatmostone(special)primewithmultiplicityatleastone.Thusif2ta(p)1isgreaterthan101g,Nhasmostthan101notnecessarilydistinctprimefactors.Sowehaveacontradiction.Forthelowerboundgofthenotnecessarilydistinctprimefactors,wecomputethesumgpoftheexponentsoftheprimesthathavebeenbranchedon,andweaddfourtimesthenumbergcofcomposites.Sincewecheckedthatacompositeisnotaperfectpower,itmustbedividedbytwodi erentprimes,eachhavingmultiplicityatleasttwo,exceptforatmostone(thespecialprime).Sowetakeg=gp+4gcorg=gp+4gc1,dependingonwhetherwehavealreadybranchedonthespecialprime.Bytheabove,wecancomputeanupperboundonthesmallestprimedividingNbutnotM.So,toprovethatthereisnooddperfectnumberwithfewerthan101notnecessarilydistinctprimefactorssuchthatMdividesN,webranchoneveryprimefactoruptothisboundtorulethemout.WedonotbranchonaprimethatdividesMorthatisalreadyforbidden.Wehavetoresorttoexactbranchingsasintheprevioussection,butthistimeonlyon32and34.WhenNhasnofactorsinS0.Weuseasuitablenotionofeciencyde nedasf0(p;a)=ln(1(pa)) a.Itistheratiobetweenthemultiplicativecontributioninabundancyandtheadditivecontributiontothenumberofprimesofthecomponentpa.Remark.ab=f0(p;a)&#x-279;&#x.055;f0(p;b).pq=f0(p;a)&#x-278;&#x.153;f0(q;a).Noticethatthebestwaytoreachabundancy2withthefewestprimesistotakecomponentswithhighesteciencyf0:Foreachallowedprimep,we ndthesmallestexponentasuchthat(pa)isnotdivisibleby4norafactorinS0.Wesortthesecomponentspabydecreasingeciencyf0togetanorderingp1;p2;p3;:::suchthatf0(p1;a1)f0(p2;a2)f0(p3;a3):::.Theproduct49=1p p1=1:99601:::issmallerthan2,whereas49=1a=118. 6PASCALOCHEMANDMICHAELRAO5.LargestcomponentofanoddperfectnumberCohen[2]provedin1987thatanoddperfectnumberhasacomponentgreaterthan1020.Theorem3.Thelargestcomponentofanoddperfectnumberisgreaterthan1062.Weusethesamealgorithmasintheprevioussectiontoforbideveryprimelessthan108usingthefollowingcontradictions:-Theabundancyofthecurrentnumberisstrictlygreaterthan2.-Thecurrentnumberhasacomponentgreaterthan1062.Sincewewanttoquicklyreachalargecomponent,webranchonthelargestavailableprime.Thereisnounfactoredcompositehere,andthusnoroadblock,sinceeverynumberislessthan1062andthushasbeeneasilyfactored.SupposenowthatNisanoddperfectnumberwithnoprimefactorlessthan108andnocomponentp�1062.First,theexponenteofanyprimefactorpislessthan8,sinceotherwisep�(108)8�1062.Theexponentofthespecialprimep1isthus1,because3j(p5)and3-N.SoNhasaprimedecompositionN=p1n2=1p2i;2n4=1p4i;4n6=1p6i;6.Let(x)denotethenumberofprimeslessthanorequaltox.Inthefollowing,wewillusetheseknownvaluesof(x)[8]:(108)=5761455(31010)=1300005926(321014)=92295556538011(981014)=273808176380030Itiswell-known(see[6])thatforprimesq,r,andssuchthatqj(rs1),eitherq=sorq1mods.Soifpj;e0j(pi;e),thenpj;e01mod(e+1),since(e+1)-N.Wethushavee0=e,sinceotherwisee+1woulddivide(pj;e)(thatis,(p0j;e0)),butnotN.Moreover,(pi;e)cannotbeprimeunlessitisthespecialprimep1.Supposetothecontrarythat(pi;e)=pj;e0.Thenp0j;e0isacomponentofN.Sincee0=e,wehavethatee08,sothatp0j;e0=((pi;e))0�(pi;e)0=(pi;e)0�(108)8�1062.Soeach(pi;e)producesatleasttwofactorsorthespecialprime.Letn2;2bethenumberofprimespi;2suchthat(p2i;2)=qrwhereqr,qandrprimes.Letn2;3bethenumberofprimespi;2suchthat(p2i;2)factorsintoatleastthreenotnecessarilydistinctprimes.Bytheabove,wehave(1)n2n2;2+n2;3+1:Bycountingthenumberofprimesproducedbythefactors(p2i;2),weobtain(2)2n2;2+3n2;34n4+6n6+1:Fore2f4;6g,wehavepi;e321014,sinceotherwisepi;e&#x-0.8;أ倀(321014)4&#x-0.8;أ倀1062.Supposethataprimepi;2issuchthat(p2i;2)=qrwhereqr,qandrprimes.Thenwehavethatr&#x-278;&#x.153;pi;2,andbypreviousdiscussion,eitherr=p1orr=p0;efore2f4;6g.Thisimpliesthatatleast(n2;21)primespi;2aresmallerthanthelargestprimepi;efore2f4;6g.So,n2;21+n4+n6(321014)(108)=92295550776556whichgives(3)n2;2+n4+n692295550776557: ODDPERFECTNUMBERSAREGREATERTHAN1015007Similarly,pi;631010sinceotherwisep6i;6&#x-0.8;أ倀1062.So,n6(31010)(108),whichgives(4)n61294244471Now,weconsideranupperboundontheabundancyofprimesgreaterthan108.Weuseequation(3.29)in[7],Ypxpprimep p1e\rln(x)1+1 2ln2(x)where\r=0:5772156649:::isEuler'sconstant.WecomputethatYp108pprimep p1�c1=32:80869860873870116andweobtainY108p981014pprimep p1e\rln(981014)1+1 2ln2(981014)=c12:Bytheabove,wehave1+n2+n4+n6&#x-0.8;͸䀀(981014)(108)=273808170618575,whichgives,(5)273808170618575n2+n4+n6:Thecombination3(1)+1(2)+7(3)+2(4)+3(5)gives6n2;2+1753530679308800,acontradiction.6.ImprovementsoverpreviousmethodsThispaperprovidesauni edframeworktoobtainlowerboundsonthreepa-rametersofanoddperfectnumber.ThemostusefulnewtoolisthewaytogetaroundroadblocksintheproofofTheorem1.Theargumenttoobtainaboundonthesmallestnot-yet-consideredprimeisanadaptationoftheonein[4].Inbothcaseitimpliesaboundb,anexponentt,aninequalityrelatedtotheabundancy,andaninequalityrelatedtothecorrespondingparameter.Theargumentismoresophisticatedinthecontextofaboundonthesizeratherthanonthetotalnumberofprimes,becausebothbandtareinvolvedinbothinequalities.Brentetal.[1]usedstandardbranchingsandHare[4]usedexactbranchings.Weintroducetheuseofacombinationofstandardandexactbranchingstoreducethesizeoftheprooftree.Standardbranchingsareeconomicalbutexactbranchingsaresometimesunavoidablewhenwehavetoby-passaroadblock.Inthe nalphaseoftheproofofTheorems1and2,wehavetoarguethatanoddperfectnumberwithnofactorsinasetofsmallforbiddenprimesnecessarilyviolatethecorrespondingbound.Whentheboundincreases,thesetofforbiddenprimesmustgetlarger.Suitablenotionsofeciencyofacomponentareintroducedinordertorestrainthegrowthofthisset.Theyallowabetteruseofthefactthatsomeprimesareforbidden,byconsideringtheexponentoftheremainingpotential 8PASCALOCHEMANDMICHAELRAO #Branch. #Branch.(circ.) Approx.time Theorem1 22514255 10406935 12hours Theorem2 447019005 444022 93hours Theorem3 6574758 0 30minutes Figure2.Totalnumberofbranchings,numberofbranchingsinroadblockscircumventing,andapproximatetime.primefactors.Finally,wegiveaproofofTheorem3usingasystemofinequalities.Theideabehindisasfollows.IfallprimesuptoBareforbidden,thenthelargestprimefactormustbeatleastB2inordertoreachabundancy2.ThenweusevariousargumentsandinequalitiesinordertoshowthatanottoosmallproportionCoftheprimefactorshaveexponentatleast4.Thenweconcludethatacomponentofsizeatleast(CB2)4=C0B8exists.Wewouldliketopointouttheimportanceofseparatingthesearchforfactors,withecientdedicatedsoftwares,fromthegenerationoftheprooftree.Inpartic-ular,thiscausesthemostpartoftheimprovementforTheorem2.7.Credit-acknowledgementTheprogramwaswritteninC++andusesGMP.Theprogramandthefactorsareavailableathttp://www.lri.fr/~ochem/opn/.WepresentinFigure2thenumberofbranchingsonprimefactors(overallandneededincircumventsofroadblocks),andthetimeneededonaAMDPhenom(tm)IIX4945toprocessthetreeofsuppositionsforeachtheorem.Ofcourse,thisdoesnottakeintoaccountthetimeneededto ndthefactors.Varioussoftwaresandalgorithmswereusedforthefactorizations:GMP-ECMforP-1,P+1andECM,msieveandyafuforMPQS,msievecombinedwithGGNFSforNFS(bothgeneralandspecial).Wethankthepeoplewhocontributedtothefactorizationsatthemersennefo-rum,yoyo@home,RSALS,andelsewhere,orprovidedhelpfulcommentsonprelimi-naryversionsofthepaper.InparticularWilliamLippwhoisobtainingandgather-ingusefulfactorizationsviahiswebsitehttp://www.oddperfect.org,TomWom-ackwhoobtainedthefactorizationof(191102)and(280178),WarutRoonguthai,CarlosPinho,ChrisK.,SergeBatalov,PaceNielsen,LionelDebroux,GregChilders,AlexanderKruppa,Je Gilchrist.ExperimentspresentedinthispaperwerepartiallycarriedoutusingthePLA-FRIMexperimentaltestbed,beingdevelopedundertheINRIAPlaFRIMdevel-opmentactionwithsupportfromLABRIandIMBandotherentities:ConseilRegionald'Aquitaine,FeDER,UniversitedeBordeauxandCNRS(seehttps://plafrim.bordeaux.inria.fr/). ODDPERFECTNUMBERSAREGREATERTHAN1015009References[1]R.P.Brent,G.L.Cohen,H.J.J.teRiele.Improvedtechniquesforlowerboundsforoddperfectnumbers,Math.Comp.(1991),no.196,pp857{868.[2]G.L.Cohen.Onthelargestcomponentofanoddperfectnumber,J.Austral.Math.Soc.Ser.A(1987),pp280{286.[3]T.Goto,Y.Ohno.Oddperfectnumbershaveaprimefactorexceeding108,Math.Comp.(2008),no.263,pp1859{1868.[4]K.G.Hare.Newtechniquesforboundsonthetotalnumberofprimefactorsofanoddperfectnumber,Math.Comp.(2007),no.260,pp2241{2248.[5]P.P.Nielsen.Oddperfectnumbershaveatleastninedi erentprimefactors,Math.Comp.(2007),no.160,pp2109{2126.[6]T.Nagell.IntroductiontoNumberTheory,JohnWiley&SonsInc.,NewYork,1951.[7]J.B.Rosser,L.Schoenfeld.Approximateformulasforsomefunctionsofprimenumbers.IllinoisJ.Math.(1962),pp64{94.[8]http://www.trnicely.net/pi/pix_0000.htmLRI,CNRS,B^at490UniversiteParis-Sud11,91405OrsaycedexFranceE-mailaddress:ochem@lri.frCNRS,LabJ.V.Poncelet,Moscow,Russia.LaBRI,351coursdelaLiberation,33405Talencecedex,FranceE-mailaddress:rao@labri.fr

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MATHEMATICS OF COMPUTATION Volume Number Pages S XX ODD PERFECT NUMBERS ARE GREATER THAN PASCAL OCHEM AND MICHA EL RAO Abstract - Description


Brent Cohen and te Riele proved in 1991 that an odd perfect number is greater than 10 300 We modify their method to obtain N 10 1500 We also obtain that has at least 101 not necessarily distinct prime factors and that its largest component ie div ID: 7338 Download Pdf

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