On values of d(n!)/m!, φ(n!)/m! and σ(n!)/m!

Transcription

1 O values of d!/m!, φ!/m! ad σ!/m! by Daiel Baczkowski, Michael Filaseta, Floria Luca ad Ogia Trifoov 1 Itroductio I [5], the third author established that for a fixed r Q, there are oly a fiite umber of positive itegers ad m for which f! = r m! where f is oe of the arithmetic fuctios d the umber of divisors fuctio, φ Euler s φ-fuctio, or σ the sum of the divisors fuctio I this paper, we establish a geeralizatio of these results A similar result that we do ot address here was give by the third author ad I Shparliski [6] for the fuctio τ, that is Ramauja s tau fuctio Deotig by ω the umber of distict prime divisors of, the followig is a easily stated cosequece of our mai results Theorem 1 Let f deote oe of the arithmetic fuctios d, φ or σ, ad let k be a fixed positive iteger The there are at most fiitely may positive itegers, m, a ad b such that b f! = a m!, gcda, b = 1 ad ωab k 1 Alteratively, Theorem 1 is assertig that the total umber of distict primes dividig the umerator ad deomiator of the fractio obtaied by reducig the quotiet f!/m! teds to ifiity as the product m teds to ifiity For the proof of Theorem 1, we ote that it suffices to show that 1 implies that there is a positive iteger N = Nk such that the iequality N holds I fact, oce N is established, we ca deduce that the left-had side of the first equatio i 1 has a bouded umber of distict prime factors depedig oly o k This the implies that m is bouded ad, hece, that there are oly a fiite umber of possibilities for the value of a/b = f!/m! Give that gcda, b = 1, we ca the deduce that there are oly a fiite umber of possibilities for the quadruple, m, a, b We will establish results cosiderably stroger tha Theorem 1 for each of the arithmetic fuctios give there Our argumet for the case f = σ will be more ivolved tha our argumets for f = d ad f = φ This is due to the difficulty i estimatig the umber of large distict prime divisors of σ! We are therefore able to more easily prove the cases whe f = d ad f = φ Theorem 2 There are at most fiitely may positive itegers a, b, ad m such that b d! = a m!, gcda, b = 1, ωb m 1/4 ad P 0 a log 22, 2 where P 0 a deotes the least prime ot dividig a 2000 Mathematics Subject Classificatio: 11A25 11N05 The secod ad fourth authors were supported by the Natioal Sciece Foudatio The secod author was also supported by the Natioal Security Agecy The third author was supported by SEP-CONACyT 46755

2 Theorem 3 There are at most fiitely may positive itegers a, b, ad m, with > 1, such that b φ! = a m!, gcda, b = 1 ad max{ωa, ωb} 7 log 3 Theorem 4 Fix ε > 0 The there are at most fiitely may positive itegers a, b, ad m such that b σ! = a m!, gcda, b = 1, ωab 02 ε 4 These three results are meat to reflect the flavor of what the argumets produce Regardig Theorem 2, the bouds o ωb ad P 0 a ca be sharpeed slightly ad altered to the extet that oe ca weake or stregthe the boud for ωb if oe wats to improve or is willig to weake, respectively the boud for P 0 a Although the boud o max{ωa, ωb} i Theorem 3 is ot sharp, by cosiderig m = /2 ad a ad b havig prime factors from [2, /2], it is ot difficult to see that oe caot, for ay ε > 0, replace the estimate max{ωa, ωb} /7 log with ωab / 2 ε log I particular, the boud is withi a costat factor of beig best possible As with Theorem 1, oe ca reduce establishig Theorem 2, 3 or 4 to showig that there is a positive iteger N for which N To see this, suppose such a N exists I the case of Theorem 2, we deduce that the umber of distict primes dividig the left-had side of the first equatio i 2 is at most a fuctio of N plus m 1/4 I the case of Theorems 3 ad 4, we deduce that the umber of distict prime factors o the left-had side of the first equatio i 3 ad 4, respectively, is bouded by a fuctio of N As the umber of distict prime divisors of m! is m/ log m by the Prime Number Theorem or simply a Chebyshev estimate, we obtai that i ay case m is bouded, ad we deduce as before that there are at most a fiite umber of quadruples, m, a, b It is ot difficult to see that Theorems 2, 3 ad 4 imply Theorem 1 for f = φ, d, ad σ, respectively To see this, let f be oe of these three multiplicative fuctios We have already see that the umber of quadruples, m, a, b as i 1 is bouded if is fixed I the case that f = d, as we will see i the ext sectio, it is also the case that the umber of quadruples is bouded if m is fixed ad, a ad b are allowed to vary For large, if ωab k, the the coditios i 3 ad 4 are satisfied For ad m large, if ωab k, the the coditios i 2 are satisfied Hece, Theorems 2, 3 ad 4 imply that there are oly a fiite umber of quadruples, m, a, b as i 1 2 The fuctio d We establish Theorem 2 ad, hece, Theorem 1 i the case that f = d Recall that it suffices to show that uder the coditios of 2, is bouded We therefore cosider large ad assume that we have a solutio, m, a, b to 2 with the goal of obtaiig a cotradictio We first show that m must also be large To establish this, we may suppose m 2 Cosider a prime q log 2 Let p be a prime i the iterval /q, /q 1] Usig the logarithmic itegral approximatio of πx i the Prime Number Theorem with a appropriate error term gives that the umber of such primes is qq 1 log 2

3 For each such prime p, we have p > so that ν p! = /p = q 1, ad we deduce that q = ν p!+1 is a prime divisor of d! Moreover, for q a prime log 2, we have that ν q d! is at least the umber of primes p /q, /q 1] Thus, ν q bd! νq d! for q log 2 5 2qq 1 log We cosider a prime q log 2 that does ot divide a, which exists by the coditio o a i 2 As is large, we deduce from 5 that the left-had side of the first equatio i 2 is divisible by a large power of q We deduce the that m too must be large Note that this implies that the umber of quadruples, m, a, b as i 2 is bouded if m is fixed as was idicated i the itroductio We will wat a geeral estimate for the expoet i the highest power of a prime p dividig! A classical formula is ν p! = Alteratively, j=1 ν p! = s p, p 1 where s p is the sum of the base p digits of cf [1] For this sectio, we use that the latter formula easily implies ν p! 1/p 1 Oe easily deduces that for a positive iteger ad p a prime, we have ν p! p p It follows that log d! = log νp! + 1 log 2 + log log p = π log 2 + log log p p p p Classical Prime Number Theorem estimates imply that the right-had side above is < 2/ log O the other had, Stirlig s formula easily gives that Ideed, this last iequality ca be see from p j logm! > m log m m e m = j=0 j m j! > mm m! For a arbitrary prime p, we use ν p m! m 1/p 1 < 2m/p Hece, log p m p b p νpm! = p m p b p m p b ν p m! log p 2m p m p b log p p If b 3, this last sum is o bigger tha the sum oe obtais with b replaced by the product of the first ωb primes This allows us to deduce from a direct computatio that if ωb 100, the log p 3 log ωb p 2 3

4 For ωb > 100, we appeal to 313 ad 323 of [8] The former implies that the ωbth prime is < 2ωb log ωb The latter implies log p < log ωb + log log ωb < 3 log ωb, p 2 p 2ωb log ωb all uder the coditio ωb > 100 I ay case, 6 holds idepedet of the value of ωb, ad we obtai log p νpm! < 3m log ωb + 1 p m p b From 2, we have log ωb log m We deduce that the logarithm of the product of the primes to their multiplicities o the right-had side of the first equatio i 2 that do ot divide b is at least m log m m 078 m log m > m log m, 5 where we have used that m is large This the is a lower boud for log d!, ad we deduce 2 log > m log m 5 As ad m are large, we obtai m 11/ log 2 Hece, ν q m! m 1 q 1 < 11 q 1 log 2 We obtai from 5 that for each q log /22, we have ν q b d! > νq m! We deduce from the first equatio i 2 that each prime log /22 divides a, cotradictig the coditio o a i Theorem 2 ad, hece, completig the proof 3 The fuctio φ I this sectio, we establish Theorem 3 ad, hece, Theorem 1 i the case that f = φ We agai cosider large ad assume that we have a solutio, m, a, b to 3 with the goal of obtaiig a cotradictio We begi by usig the boud /7 log o ωb Observe that every prime factor of φ! is /2 Also, usig that is large ad the Prime Number Theorem, we have π065 π05 > 7 log ωb It follows that there is a prime 065 that does ot divide b φ! As every prime m divides the right-had side of the first equatio i 3, we deduce m < 065 Next, we use that ωa /7 log This iequality together with beig large ad the Prime Number Theorem imply that there is a prime q satisfyig < q 015 that does ot divide a We fix such a q Observe that the expoet i the largest power of q dividig φ! is at least /q 1 > /q 2 Sice m < so that q > > m, the expoet i the largest 4

5 power of q dividig m! is m/q m/q As q a, we have a cotradictio to the first equatio i 3 if q 2 m q Give that m < 065 ad q 015, we have The theorem follows q m q > 035 q 4 Prelimiaries for the fuctio σ I this sectio, we discuss a few prelimiary results we will use i our proof of Theorem 1 for f = σ We deote by Φ N x the Nth cyclotomic polyomial We use the followig cyclotomic polyomial idetities Let N be a positive iteger, ad let p be a prime The { Φ N x p if p N Φ pn x = 7 Φ N x p /Φ N x if p N > 2 I additio, we use that x N 1 = d N Φ d x 8 We begi by aalyzig the highest power of a give prime q that ca divide a expressio of the form a N 1 We have i mid here obtaiig iformatio about the prime factorizatio of σ! which ivolves factors of the form p N 1/p 1 The approach we use takes advatage of the factorizatio give i 8; hece, we are iterested i estimates for ν q Φd a We use the otatio ord q a for the order of a modulo q The ext result, at least for the most part, is fairly well-kow We give a proof that is motivated i part by a 1905 paper by L E Dickso [2] Lemma 1 Let q be a prime, ad let a ad N be itegers with N 1 Write N = q r M where r ad M are itegers with r 0 ad q M The q Φ N a if ad oly if M = ord q a Also, if r 1 ad N > 2, the q 2 Φ N a Proof To prove the first assertio, let s = ord q a First, cosider the case that M = s We obtai from 8 that Φ d a a M 1 0 mod q d M If q Φ d a for some d, the a d 1 mod q so that M d sice M = ord q a We deduce that the oly factor o the left that ca be divisible by q is Φ M a Hece, q Φ M a Observe that 7 implies Φ N x Φ M x qr 1 q 1 mod q 9 Settig x = a, we deduce q Φ N a Assume ow that q Φ N a ad M s We wat to obtai a cotradictio Sice q Φ N a, we have a M a N 1 mod q We deduce a 0 mod q, s M ad, hece, s < M It follows 5

6 that x s 1Φ M x is a factor of x M 1 The defiitio of s implies x a is a factor of x s 1 modulo q From 9 ad q Φ N a, we have that x a is a factor of Φ M x modulo q We obtai that x M 1 x a 2 gx mod q for some gx Z[x] Oe obtais a cotradictio by takig derivatives ad settig x = a For the secod assertio, we use that Φ N x is a factor of x N/q q 1 = x N/q q 1 + x N/q q x N/q 2 + x N/q + 1 x N/q 1 Substitutig x = a o the left, we see that if q Φ N a, the ecessarily a N/q 1 mod q Writig a N/q = kq + 1, where k Z, observe that the expressio o the right with x N/q replaced by a N/q is kq + 1 q 1 + kq + 1 q kq q + kq2 q 1 mod q 2 2 Sice the left-had side is divisible by Φ N a, we see that if q 2, the q 2 Φ N a, as desired If q = 2, i fact a stroger assertio is true I this case, q 2 Φ N a idepedet of whether q N To see this, ote that for every prime p, we have Φ p 1 = p From 7, Φ N 1 = p if N is a power of a prime p ad Φ N 1 = 1 if N is a iteger with more tha oe distict prime factor Also, N > 1 implies Φ N 0 = 1 Hece, Φ N a 1 mod 2 if N is ot a power of 2 or if a is eve If N = 2 r with r 2 ad if a is odd, the Thus, i ay case, 4 Φ N a for N > 2 Φ N a Φ 2 ra a 2r mod 4 The followig cosequece of Lemma 1 is worth otig Corollary 1 Let q be a prime, ad let a ad N be itegers with N > 2 If q Φ N a, the either q 1 mod N or we have that both q is the largest prime factor of N ad q 2 Φ N a The coditio N > 2 i each of the above results is importat as Φ 2 a = a + 1 ca clearly, for the right choice of a, be divisible by a arbitrarily large power of 2 Lemma 2 Let q be a odd prime, ad let j ad l be positive itegers Let fx = x l + x l x + 1 The fx has gcd φq j, l + 1 distict roots modulo q j Furthermore, fx has 2 gcd φ2 j, l + 1 distict roots modulo 2 j Proof Observe that x 1fx = x l+1 1 Let N = l + 1 The lemma follows from the fact cf [4], page 45 that x N 1 mod q j has exactly gcd φq j, N roots modulo q j if q is odd or q = 2 ad j {1, 2} ad has exactly 1 or 2 gcd 2 j 2, N roots modulo 2 j if j 3 depedig o whether N is odd or eve, respectively I regards to Lemma 2, we will oly be usig that the umber of distict roots of fx modulo q j is gcd φq j, l + 1, ad we could easily get away with oly cocerig ourselves with odd primes q The above lemma will, however, allow us ot to worry about the parity of q i our lemmas We will also make use of the followig versio of the Bru-Titchmarsh iequality cf [7] 6

7 Lemma 3 Let x ad y be positive real umbers, ad let h ad k be itegers with 1 k < y x The umber of primes i the iterval x, x + y] that are h modulo k is 5 The fuctio σ 2y φk logy/k I this sectio, we establish Theorem 1 i the case that f = σ To avoid cofusio with our use of d whe talkig about divisors of ad the umber of divisors fuctio d discussed earlier i this paper, we use here the otatio σ 0 ie, the sum of the divisors of each raised to the power 0 for the umber of divisors of We also use the otatios f g ad f g to deote f 1 + o1 g ad f 1 + o1 g, respectively All asymptotic estimates i this sectio usig or will be with respect to Lemma 4 Let q be a prime, ad let a ad N be itegers with a > 1 ad N > 0 The ν q a N 1 log N + ord qa log a + loga + 1, where the term loga + 1 i the umerator is oly ecessary i the case q = 2 Proof To obtai our result, we estimate the power of q dividig each factor o the right of a N 1 = d N Φ da Writig N = q r M where r is a oegative iteger ad M is a positive iteger relatively prime to q, we observe that each divisor of N ca be writte uiquely i the form q j d where j is a oegative iteger r ad d is a divisor of M I other words, r a N 1 = Φ q j da j=0 d M Settig d = ord q a, Lemma 1 implies that the oly factors o the right that are divisible by q are of the form Φ q j d a We cosider the factors o the right with j 1 ad j = 0 separately Observe that r log N We apply Lemma 1 to Φ q j da, otig that the case q j d 2 below requires separate cosideratio We obtai r r ν q Φ q j da ν q Φ q j d a loga j=1 j=1 Also, The lemma follows d M r + loga + 1 ν q Φ d a = ν q Φd a logad 1 d M 7 log N + loga + 1 < d log a

8 Lemma 5 Let q be a prime umber, j a positive iteger, ad L 1 The l L gcdφq j, l l 2 log logq + 1 Before goig to the proof, we make some observatios First, the wordig of the above lemma is somewhat awkward but appropriate for our eeds A cleaer statemet would be to take the sum o the left to be a ifiite series ad the to assert that the series coverges ad its value is O log logq + 1 The implied costat here ad i the lemma is absolute The reaso for usig log logq + 1 istead of log is simply to hadle the case q = 2 where log is egative Proof Notice that for a positive iteger r, l L gcdr, l l 2 = t r l L gcdr,l=t t l = 2 t r s L/t gcdr/t,s=1 Whe r = φq j = q j 1 q 1, the right-had side above is sice σn N log log N σq 1 < ζ2 q 1 i=0 1 s 2 t ζ2 t r 1 log logq + 1 qi Lemma 6 Let r be a positive iteger, L 1, ad M = mi{r, L} The Moreover, if K 1, the l q 2 l L l gcdr, l L 2 d M d r φd d l gcdφq K, l 3q 4 2 σ 0 q 1 + 9, 1 t = ζ2σr r where q is a prime ad deotes the logarithm of to the base q ie, = log / Proof Usig the fact that d N φd = N, we have l gcdr, l = l φd = lφd l L l L d gcdr,l d M l L d r d l d<q3 d q 2 q 1 = d M d r t L/d tdφd L 2 d M d r φd d Now, let r = φq K = q K 1 q 1 ad L = q 2 If q >, the the sum o the right-had side above is φd σ 0 q 2 q 1 3σ 0 q 1 d 8

9 I the case that q, we easily have l gcdr, l l 2 L 3 = q l L l L The lemma follows Lemma 7 If 0 < ɛ < 1/5 ad q is a prime 1/5 ɛ, the log logq + 1 i ν q σ! q log If 0 < δ < 1/3 ad q is a prime, the ii ν q 1 δ <p σ p νp! log logq + 1 q + 3δ log Proof Let ep = ν p!, ad set Np = ep + 1 We also let L = q 2 We begi by provig the first part of the lemma We will estimate the cotributio of factors of q 1/5 ɛ arisig from σ p ep separately depedig o whether p /L or p > /L I other words, otig that σ! = p /L we combie estimates for ad σp ep /L<p V = Vq = ν q V = V q = ν q p /L /L<p From Lemma 4, for ay prime q we have that V p /L σp ep = p /L p Np 1 p 1 σp ep = ν q p /L σp ep = ν q /L<p log Np + ord q p log p, /L<p p Np 1 p 1 p Np 1 p 1 p Np 1 p 1, where the term loga + 1 appearig i the umerator of the boud give i Lemma 4 has bee absorbed by the implied costat above We use that Np, which follows easily from ep = < u=1 p u u=1 p u = p 1 ad that ord q p divides φq Sice also πx x/ log x, we obtai V p /L log + 9 p /L q log p, 10

10 π /L log + q q L q log p /L We divide up our cosideratio of larger primes p as follows For each positive iteger l < L, we cosider the cotributio of q s from σ p ep with p I l = /l + 1, /l] Fix such a l ad a prime p I l As is sufficietly large, the defiitio of L implies p > Sice p I l, we obtai Np = /p + 1 = l + 1 Let f l x = x l + x l x 2 + x + 1 The σp ep = f l p Observe that this polyomial defiig σp ep does ot chage as p varies over the primes i I l We ow let p vary over the primes i I l ad use that ν q σ = ν q fl p = 1 11 p I l j 1 p I l p ep log p p I l f l p 0 mod q j Let J = log + 1 so that q J = q log For each l < L, we obtai from q 1/5 ɛ that We cosider the umbers I l q J L 2 q J q 5 2 log 5ɛ log 3 2 ɛ 12 ρ j,l = ρ j,l q = {t Z : 0 t q j 1, f l t 0 mod q j } For each a {0, 1,, q j 1} with j J ad f l a 0 mod q j, we obtai from Lemma 3 that π /l; q j, a π /l + 1; q j, a 2 I l φq j log I l /q j Now, we cosider the j > J Recall that Np = l + 1 for each p I l Defie K = K l = l + 1 We show that K ca be used as a upper boud o the j appearig i 11 as follows Observe that if q j divides f l p for some p I l, the N = l + 1 implies that q j p N 1/p 1 We deduce that q j < p N ad, hece, j < N p N = l + 1 = K 13 Thus, for l fixed, we eed oly cosider the positive iteger values of j such that J < j < K For each such j, we proceed as before by coutig the umber of primes p I l such that q j divides f l p For each j > J, we simply use that the umber of primes p I l for which q j divides f l p is Il ρ j,l + 1 q j 10

11 Altogether, we deduce that V = ν q fl p l<l p I l l<l 1 j J 2 I l ρ j,l φq j log I l /q j + I l ρ j,l + q j J<j<K L J<j<K l ρ j,l We view the right-had side above as three double sums ad estimate each i tur Note that I l /l ad, from Lemma 2, we have ρ j,l 2 gcd φq j, l + 1 From the estimate i 12 ad Lemma 5, we deduce l<l 1 j J Recall that q J = q log Hece, l<l 2 I l ρ j,l φq j log I l /q j 1 j J l<l 1 j J l<l 1 j J j=1 I l ρ j,l q j J<j<K L gcd φq j, l + 1 φq j l log I l /q j gcd φq j, l + 1 φq j l log log logq + 1 φq j log log logq + 1 q j log log logq + 1 q log J<j<K L l<l j>j Recall K l = l + 1, ad observe that From Lemma 6, we obtai ρ j,l J<j<K l l<l gcd φq j, l + 1 log logq + 1 q j log logq + 1 q J log logq + 1 q log q j l ρ j,l 2 gcd φq j, l gcd φ q K l, l + 1 l<l l<l J<j<K l gcd φ q Kl, l + 1 K l gcd φ q K l, l

12 l<l l + 1 gcd φ q K l, l + 1 q 4 σ 0 q For fixed ɛ > 0 ad q 1/5 ɛ, this sum is 1 ɛ /q /q log Combiig the above, we obtai for q 1/5 ɛ that ν q σ! = V + V log logq + 1 q log For the secod part of the lemma, we ca give a similar but simpler argumet We take L = δ We partitio the iterval I l ito cogruece classes of legth q j For each l < L, we cosider all possible values of 1 j K l together Doig so, we obtai ν q fl p p I l l<l ρ j,l l<l 1 j<k l Il q j j<k L l<l ρ j,l I l q j + ρ j,l l<l 1 j<k L Applyig Lemma 2 ad Lemma 5 to the first double sum o the right-had side above ad usig that ρ j,l l to the latter, we obtai p Np 1 ν q = ν q fl p log logq δ log p 1 q l< δ p I l 1 δ <p The lemma follows Proof of Theorem 4: Set c = 1/5 2ɛ where 0 < ɛ < 1/10 It suffices to show that ωab c has o solutios for sufficietly large So assume is sufficietly large ad ωab c First, we cosider the case that ω σ! 2 c The there exists c distict primes p dividig σ! ad ot dividig ab The equatio b σ! = a m! implies that ay such prime p must divide m! ad, hece, every prime p divides b σ! We deduce that amog the first 2 c primes, there is a odd prime q that divides σ! ad ot ab Note that q c+ɛ 1/5 ɛ Sice q does ot divide ab, we have Lemma 7 i ow implies ν q σ! = νq m! m q 1 m log log 14 log Before proceedig, we ote that the case whe ω σ! < 2 c also gives 14 Ideed, i this case we have m log m πm = ωm! ω b σ! ωb + ω σ! c implyig that m c log Observe that log σ! log! log 12

13 ad, from 14, Hece, b σ! = a m! implies Fix 0 < δ < 1/3 Also, let a = logm! m log m log log log a = logb/m! + log σ! log p 1 δ σ Clearly, a a a As a cosequece of 10, we have from which we deduce Combiig the above, we get p ep ad a = gcd a, σ!/a p 1 < ep < p 1 log a ep log p 1 δ log p 1 δ log log a log a + log a 1 δ log + q a ν q a 15 From Lemma 7, q a ν q a q a q c+ɛ log log q log + log logq δ log q q a q> c+ɛ For the first sum o the right, we have q a q c+ɛ log log q log log log log q q c+ɛ log log For the secod sum, we use that the umber of terms is bouded by ωa We obtai ad q a q> c+ɛ log logq + 1 q q a q> c+ɛ ωa log logc+ɛ + 1 c+ɛ c log log c+ɛ 3δ log ωa 3δ log 13 log c+ɛ log 1 ɛ/2

14 Thus, q a ν q a log log + 1 ɛ/2 + ωa 3δ log By 15, this last sum must exceed 1 + o1δ log Cosequetly, ωa 3δ log log Takig δ = 4/15 < 1/3, the left-had side is ad we reach the desired cotradictio Hece, the theorem is complete Refereces [1] G Bachma, Itroductio of p-adic umbers ad valuatio theory, Academic Press, New York, 1964 [2] L E Dickso, O the cyclotomic fuctio, Amer Math Mothly 12, No , [3] G H Hardy ad E M Wright, A Itroductio to the Theory of Numbers, Fifth editio, Oxford Uiv Press, Oxford, 1989 [4] K Irelad ad M Rose, A Classical Itroductio to Moder Number Theory, Secod editio, Spriger-Verlag, New York, 1990 [5] F Luca, Equatios ivolvig arithmetic fuctios of factorials, Divulg Mat , [6] F Luca ad I Shparliski, Arithmetic Properties of the Ramauja Fuctio, Proceedigs of the Idia Academy of Scieces Mathematical Scieces, , 1 8 [7] H L Motgomery ad R C Vaugha, The large sieve, Mathematika , [8] J B Rosser ad L Schoefeld, Approximate formulas for some fuctios of prime umbers, Illiois J Math ,