143K - views

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

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

Tags : Brent Cohen and
Embed :
Pdf Download Link

Download Pdf - The PPT/PDF document "MATHEMATICS OF COMPUTATION Volume Numbe..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.

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






Presentation on theme: "MATHEMATICS OF COMPUTATION Volume Number Pages S XX ODD PERFECT NUMBERS ARE GREATER THAN PASCAL OCHEM AND MICHA EL RAO Abstract"— Presentation transcript:

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