#A81 INTEGERS 13 (2013) THE AVERAGE LARGEST PRIME FACTOR
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1 #A8 INTEGERS 3 (03) THE AVERAGE LARGEST PRIME FACTOR Eric Naslund Dearmen of Mahemaics, Princeon Universiy, Princeon, New Jersey naslund@mahrinceonedu Received: /8/3, Revised: 7/7/3, Acceed:/5/3, Published: /3/3 Absrac Le P (n) denoe he larges rime facor of an ineger n In his aer we look a he average of P (n), and rove ha P (n) = li g () + O e c(log )3/5, nale where li g () = R [/] log d allows us o deduce heir asymoic eansion P (n) = c 0 log + c (log ) + + c k nale This imroves a resul of De Koninck and Ivíc and (k ) (log ) k + O (log ) k+ as a corollary, wih he advanage ha we can give he consans elicily as c n = n j ( ) j (j) () n+ Inroducion Le P (n) denoe he larges rime facor of an ineger n The momens of P (n) were firs looked a in 976 by Knuh and Pardon [4], and concerning he average, hey roved ha nale nale P (n) log () The above asymoic also aeared in Erdős and Alladi s 977 aer [] on addiive arihmeic funcions Laer, De Koninck and Ivić [] roved ha P nale P (n) has an asymoic eansion of he form P (n) = d log + d log + + d m log m + O log m+
2 INTEGERS: 3 (03) where he consans d i are comuable, bu no given eliciely This eansion aears again in [3], where Ivić finds a similar formula for he k h larges rime facor In his aer, we calculae he average of P (n) u o an error of he form O e c(log )3/5, like ha of he rime number heorem This allows us o deduce De Koninck and Ivić s eansion as a corollary, as well as give he consans d i eliciely Our main resul is: Theorem Leing li g () = R nale [/] log d, we have ha P (n) = li g () + O e c(log )3/5 () For any ineger k, li g () has he asymoic eansion li g () = c 0 log + c (log ) + + c (k ) k (log ) k + O (log ) k+ where c n = n+ n These consans c n saisfy c n = + O n n Noice in aricular ha c 0 =, so we deduce ha nale (3) j ( ) j (j) () (4) P (n) = log + O (log ) Alhough his eansion has reviously been resened in [], and again in [3], here are some advanages o he resul above We are able o give he consans c n eliciely in erms of he zea funcion wihou much addiional work, and our roof is basic requiring only an alicaion of he hyerbola mehod Furhermore, we give he elici inegral form which has an error erm like ha of he rime number heorem, aralleling how () is aroimaed by he funcion li() The Main Theorem For each ineger n ale, here is a mos one rime > h i such ha n, and for each rime here will be eacly inegers less han which are divisible by Combining hese wo facs, we see ha ale <ale ale nale P (n) ale ale ale,
3 INTEGERS: 3 (03) 3 and since P ale h h By wriing i i ale P ale = O nale P (n) = ale ale 3/ log, i follows ha 3/ + O (5) log = P nale: n, and rearranging he order of summaion, we obain ale = = (6) ale ale nale: n nale Leing K() = P ale, we have he esimae Z K() = log d + O e c(log )3/5, (7) which follows from he rime number heorem (See [5]) along wih arial summaion Using (7) along wih equaions (5), and (6), we have ha P (n) = Z n nale nale log d + O n e c(log n) 3/5 + O 3 log nale Z = log d + O e c(log )3/5 (8) ale n = li g () + O e c(log )3/5, (9) which roves () To recover he asymoic eansion in (3), we urn our aenion o his inegral funcion and make he subsiuion = u o obain li g () = Z Z / log d = du (0) We can rewrie he inegral above as Z / and for any ineger k log we have ha Z / u du = log = + log + + u Z / u 3, by he geomeric series eansion k + log log du = k Z / (log ) n+ + (log ) k+ Z / u du, log k u 3 ()n du ()k u3, log du log
4 INTEGERS: 3 (03) 4 Each of hese inegrals is absoluely convergen on [, ), and so he las erm conribues an error erm of a mos O k Seing c R log k+ k = k u () k du, 3 and alying he bound Z Z u 3 (log () n n (log ) u)n du ale u du = O, / / i follows ha li g () = k nc n log (log ) n + O k log k+ () To evaluae hese consans c n, we use eonenial generaing series Inegraion by ars ells us ha for s > Z Z (s) = n s = s d [] = s [] s d, and so n= c k z k = k=0 k=0 Z [] z 3 d = ( z) z Mulilying he ower series eansions for ( z) and z yields 0 ( ) j (j) () zj A z k k = z ( ) j (j) () A k, and hence c n = n+ To find he size of c n we noe ha ( ) j (j) () = (log k) j k = k= Z (log ) j n d + k+j=n j ( ) j (j) () Z (j log ) log j 3 {} d Subsiuing = e u, he firs inegral becomes (j + ), and he second is O, j which imlies ha ( ) j (j) () = O j, and so c n = + O n n This esablishes all of Theorem Acknowledgmen I am graeful o Greg Marin for his suggesions and encouragemen
5 INTEGERS: 3 (03) 5 References [] K Alladi and P Erdős On an addiive arihmeic funcion Pacific J Mah 7():75 94, 977 [] Jean-Marie De Koninck and Aleksandar Ivić The disribuion of he average rime divisor of an ineger Arch Mah (Basel) 43() (984), [3] Aleksandar Ivić On he kh rime facor of an ineger Zb Rad Prirod-Ma Fak Ser Ma 0() (990), [4] Donald E Knuh and Luis Trabb Pardo Analysis of a simle facorizaion algorihm Theore Comu Sci 3(3) (976/77), [5] Hugh L Mongomery and Rober C Vaughan Mulilicaive number heory I Classical heory, volume 97 of Cambridge Sudies in Advanced Mahemaics Cambridge Universiy Press, Cambridge, 007
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