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2. Preliminary results
Let (α, β) = (1+ √2, 1 √2) be the roots of the characteristic equation x2 2x 1 = 0 of the Pell sequence− P . The Binet formula for P is − − { n}n≥0 n αn βn (2) P = − for all n 0. n α β ≥ − This implies easily that the inequalities (3) αn−2 P αn−1 ≤ n ≤ hold for all positive integers n.
We let Qn n≥0 be the companion Lucas sequence of the Pell sequence given by Q = 2, Q{ =} 2 and Q =2Q + Q for all n 0. Its first few terms are 0 1 n+2 n+1 n ≥ 2, 2, 6, 14, 34, 82, 198, 478, 1154, 2786, 6726, 16238, 39202, 94642, 228486, 551614,...
The Binet formula for Qn is (4) Q = αn + βn for all n 0. n ≥ We use the well-known result. Lemma 2. The relations
(i) P2n = PnQn, (ii) Q2 8P 2 = 4( 1)n n − n − hold for all n 0. ≥
For a prime p and a nonzero integer m let νp(m) be the exponent with which p appears in the prime factorization of m. The following result is well-known and easy to prove. Lemma 3. The relations
(i) ν2(Qn)=1, (ii) ν2(Pn)= ν2(n) hold for all positive integers n.
The following divisibility relations among the Pell numbers are well-known. Lemma 4. Let m and n be positive integers. We have: (i) If m n then P P , | m | n (ii) gcd(Pm, Pn)= Pgcd(m,n).
For each positive integer n, let z(n) be the smallest positive integer k such that n Pk. It is known that this exists and n Pm if and only if z(n) m. This number is| referred to as the order of appearance of| n in the Pell sequence.| Clearly, z(2) = 2. 2 Further, putting for an odd prime p, e = , where the above notation stands p p for the Legendre symbol of 2 with respect top, we have that z(p) p e . A prime | − p factor p of Pn such that z(p) = n is called primitive for Pn. It is known that Pn has a primitive divisor for all n 2 (see [2] or [1]). Write P = pep m , where ≥ z(p) p PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 3
k mp is coprime to p. It is known that if p Pn for some k>ep, then pz(p) n. In particular, | | (5) ν (P ) e whenever p ∤ n. p n ≤ p We need a bound on ep. We have the following result. Lemma 5. The inequality (p + 1) log α (6) e . p ≤ 2 log p holds for all primes p.
Proof. Since e2 = 1, the inequality holds for the prime 2. Assume that p is odd. Then z(p) p + ε for some ε 1 . Furthermore, by Lemmas 2 and 4, we have | ∈ {± } pep P P = P Q . | z(p) | p+ε (p+ε)/2 (p+ε)/2 By Lemma 2, it follows easily that p cannot divide both Pn and Qn for n = (p+ε)/2 since otherwise p will also divide Q2 8P 2 = 4, n − n ± ep a contradiction since p is odd. Hence, p divides one of P(p+ε)/2 or Q(p+ε)/2. If ep p divides P(p+ε)/2, we have, by (3), that pep P P < α(p+1)/2, ≤ (p+ε)/2 ≤ (p+1)/2 which leads to the desired inequality (6) upon taking logarithms of both sides. In ep case p divides Q(p+ε)/2, we use the fact that Q(p+ε)/2 is even by Lemma 3 (i). ep Hence, p divides Q(p+ε)/2/2, therefore, by formula (4), we have (p+1)/2 Q(p+ε)/2 Q(p+1)/2 α +1 pep < < α(p+1)/2, ≤ 2 ≤ 2 2 which leads again to the desired conclusion by taking logarithms of both sides. ⊓⊔ For a positive real number x we use log x for the natural logarithm of x. We need some inequalities from the prime number theory. For a positive integer n we write ω(n) for the number of distinct prime factors of n. The following inequalities (i), (ii) and (iii) are inequalities (3.13), (3.29) and (3.41) in [15], while (iv) is Th´eor´eme 13 from [6]. Lemma 6. Let p
1.79 log log n +2.5/ log log n holds for all n 3. ≥ 4 BERNADETTE FAYE AND FLORIAN LUCA
(iv) The inequality log n ω(n) < log log n 1.1714 − holds for all n 26. ≥
For a positive integer n, we put n = p : z(p) = n . We need the following result. P { } Lemma 7. Put 1 S := . n p 1 pX∈Pn − For n> 2, we have 2 log n 4 + 4 log log n (7) S < min , . n n φ(n)
Proof. Since n > 2, it follows that every prime factor p n is odd and satisfies the congruence p 1 (mod n). Further, putting ℓ := #∈ P , we have ≡± n Pn (n 1)ℓn p P < αn−1 − ≤ ≤ n pY∈Pn (by inequality (3)), giving (n 1) log α (8) ℓ − . n ≤ log(n 1) − Thus, the inequality n log α (9) ℓ < n log n holds for all n 3, since it follows from (8) for n 4 via the fact that the function x x/ log x is≥ increasing for x 3, while for n =≥ 3 it can be checked directly. To prove7→ the first bound, we use (9)≥ to deduce that 1 1 S + n ≤ nℓ 2 nℓ 1≤Xℓ≤ℓn − 2 1 1 1 + ≤ n ℓ m 2 − m 1≤Xℓ≤ℓn mX≥n − 2 ℓn dt 1 1 +1 + + ≤ n t n 2 n 1 Z1 ! − − 2 n log ℓ +1+ ≤ n n n 2 − 2 (log α)e2+2/(n−2) (10) log n . ≤ n log n Since the inequality log n> (log α)e2+2/(n−2) holds for all n 800, (10) implies that ≥ 2 log n S < for n 800. n n ≥ PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 5
The remaining range for n can be checked on an individual basis. For the second bound on Sn, we follow the argument from [10] and split the primes in n in three groups: P (i) p< 3n; (ii) p (3n,n2); (iii) p>n∈ 2; We have (11) 1 1 1 1 1 10.1 1 + + + + < , n 0 (mod 2), T = n 2 n 2n 2 2n 3n 2 3n ≡ 1 p 1 ≤ − 1 − 1 − 7.1 p∈Pn − + < , n 1 (mod 2), p
log d log p n < . d p 1 φ(n) Xd|n Xp|n − 6 BERNADETTE FAYE AND FLORIAN LUCA
Throughout the rest of this paper we use p, q, r with or without subscripts to denote prime numbers.
3. Proof of The Theorem
3.1. Some lower bounds on m and ω(Pn). We start with a computation showing that there are no other solutions than n = 1, 2 when n 100. So, from now on n> 100. We write ≤
α1 αk (15) Pn = q1 ...qk , where q < < q are primes and α ,...,α are positive integers. Clearly, m
Case 1: n is odd.
Here, we have ℓ 1. So, ≥ P n > α 10−40 > 2.4142. Pm −
Since n is odd, it follows that Pn is divisible only by primes q such that z(q) is odd. Among the first 10000 primes, there are precisely 2907 of them with this property. They are = 5, 13, 29, 37, 53, 61, 101, 109,..., 104597, 104677, 104693, 104701, 104717 . F1 { } Since k 1 −1 P 1 −1 1 < 1.963 < 2.4142 < n = 1 , − p Pm − qi 1 i=1 pY∈F Y PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 7 we get that k> 2907. Since 2k φ(P ) P , we get, by Lemma 3, that | n | m (18) n>m> 22907.
Case 2: n 2 (mod 4). ≡ Since both m and n are even, we get ℓ 2. Thus, ≥ P (19) n > α2 10−40 > 5.8284. Pm −
If q is a prime factor of Pn, as in Case 1, we have that z(q) is not divisible by 4. Among the first 10000 primes, there are precisely 5815 of them with this property. They are = 2, 5, 7, 13, 23, 29, 31, 37, 41, 47, 53, 61,..., 104693, 104701, 104711, 104717 . F2 { } Writing p as the jth prime number in , we check with Mathematica that j F2 415 1 −1 1 = 5.82753 ... − p i=1 i Y 416 1 −1 1 = 5.82861 ..., − p i=1 i Y which via inequality (19) shows that k 416. Of the k prime factors of P , we ≥ n have that only k 1 of them are odd (q1 = 2 because n is even), but one of those − k is congruent to 1 modulo 4 by McDaniel’s result. Hence, 2 φ(Pn) Pm, which shows, via Lemma 3, that | | (20) n>m 2416. ≥ Case 3: 4 n. | In this case, since both m and n are multiples of 4, we get that ℓ 4. Therefore, ≥ P n > α4 10−40 > 33.97. Pm − Letting p
2000. Since 2k φ(P )= P , we get | n m (21) n>m 22000. ≥ To summarize, from (18), (20) and (21), we get the following results. Lemma 9. If n> 2, then
(1) 2k m; (2) k | 416; (3) n>m≥ 2416. ≥ 8 BERNADETTE FAYE AND FLORIAN LUCA
3.2. Bounding ℓ in term of n. We saw in the preceding section that k 416. Since n>m 2k, we have ≥ ≥ log n (22) k < k(n) := . log 2
Let pj be the jth prime number. Lemma 6 shows that p p k(n)(log k(n) + log log k(n)) := q(n). k ≤ ⌊k(n)⌋ ≤ We then have, using Lemma 6 (ii), that
k P 1 1 1 m = 1 1 > . P − q ≥ − p 1.79 log q(n)(1 + 1/(2(log q(n))2)) n i=1 i Y 2≤pY≤q(n) Inequality (ii) of Lemma 6 requires that x 286, which holds for us with x = q(n) because k(n) 416. Hence, we get ≥ ≥ 1 P 1 1.79 log q(n) 1+ > n > αℓ 10−40 > αℓ 1 . (2(log q(n))2) P − − 1040 m Since k 416, we have q(n) > 3256. Hence, we get ≥ 1 −1 1 log q(n) 1.79 1 1+ > αℓ, − 1040 2(log(3256))2 ! which yields, after taking logarithms, to log log q(n) (23) ℓ +0.67. ≤ log α The inequality (24) q(n) < (log n)1.45 holds in our range for n (in fact, it holds for all n > 1083, which is our case since for us n> 2416 > 10125). Inserting inequality (24) into (23), we get log log(log n)1.45 log log log n ℓ< +0.67 < +1.1. log α log α Thus, we proved the following result. Lemma 10. If n> 2, then log log log n (25) ℓ< +1.1. log α
3.3. Bounding the primes qi for i = 1,...,k. Write (26) P = q q B, where B = qα1−1 qαk−1. n 1 ··· k 1 ··· k Clearly, B φ(P ), therefore B P . Since also B P , we have, by Lemma 4, that | n | m | n B gcd(Pn, Pm)= Pgcd(n,m) Pℓ where the last relation follows again by Lemma 4 because| gcd(n,m) ℓ. Using| the inequality (3) and Lemma 10, we get | (27) B P αn−m−1 α0.1 log log n. ≤ n−m ≤ ≤ PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 9
To bound the primes qi for all i = 1,...,k, we use the inductive argument from Section 3.3 in [11]. We write k 1 φ(P ) P 1 = n = m . − q P P i=1 i n n Y Therefore, k 1 P P P P P P 1 1 =1 m = n − m n − n−1 > n−1 . − − q − P P ≥ P P i=1 i n n n n Y Using the inequality (28) 1 (1 x ) (1 x ) x + +x valid for all x [0, 1] for i =1, . . . , s, − − 1 ··· − s ≤ 1 ··· s i ∈ we get, P k 1 k 1 k n−1 < 1 1 < , P − − q ≤ q q1 n i=1 i i=1 i Y X therefore, P (29) q < k n < 3k. 1 P n−1 Using the method of the proof of inequality (13) in [11], one proves by induction on the index i 1,...,k that if we put ∈{ } i ui := qj , j=1 Y then i 2.1 (3 −1)/2 (30) ui < 2α k log log n . In particular, k q q = u < (2α2.1k log log n)(3 −1)/2, 1 ··· k k which together with formula (23) and (27) gives
k k P = q q B < (2α2.1k log log n)1+(3 −1)/2 = (2α2.1k log log n)(3 +1)/2. n 1 ··· k n−2 Since Pn > α by inequality (3), we get (3k + 1) (n 2) log α< log(2α2.1k log log n). − 2 Since k< log n/ log 2 (see (22)), we get 2 log α 3k > (n 2) 1 − log(2α2.1(log n)(log log n)(log 2)−1) − n > 0.17(n 2) 1 > , − − 6 where the last two inequalities above hold because n> 2416. So, we proved the following result. Lemma 11. If n> 2, then 3k > n/6. 10 BERNADETTE FAYE AND FLORIAN LUCA
3.4. The case when n is odd. Assume that n> 2 is odd and let q be any prime factor of Pn. Reducing relation (31) Q2 8P 2 = 4( 1)n n − n − 2 of Lemma 2 (ii) modulo q, we get Qn 4 (mod q). Since q is odd, (because n is odd), we get that q 1 (mod 4). This≡− is true for all prime factors q of P . Hence, ≡ n k 4k (q 1) φ(P ) P , | i − | n | m i=1 Y which, by Lemma 3 (ii), gives 4k m. Thus, | n>m 4k, ≥ inequality which together with Lemma 11 gives
log 4/ log 3 n log 4/ log 3 n> 3k > , 6 so n< 6log 4/ log(4/3) < 5621, in contradiction with Lemma 9.
3.5. Bounding n. From now on, n> 2 is even. We write it as n =2srλ1 rλt =: 2sn , 1 ··· t 1 where s 1, t 0 and 3 r1 < < rt are odd primes. Thus, by inequality (17), we have ≥ ≥ ≤ ··· 1 1 P αℓ 1 < αℓ < n − 1040 − 1040 φ(P ) n 1 = 1+ p 1 pY|Pn − 1 = 2 1+ , p 1 d≥3 p∈Pd − Yd|n Y and taking logarithms we get 1 1 ℓ log α < log αℓ 1 − 1039 − 1040 1 < log 2 + log 1+ p 1 d≥3 p∈Pd − Xd|n X
(32) < log 2 + Sd. d≥3 Xd|n In the above, we used the inequality log(1 x) > 10x valid for all x (0, 1/2) with x =1/1040 and the inequality log(1 + x−) x valid− for all real numbers∈ x with x = p for all p and all divisors d n with≤d 3. ∈Pd | ≥ PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 11
Let us deduce that the case t = 0 is impossible. Indeed, if this were so, then n is a power of 2 and so, by Lemma 9, both m and n are divisible by 2416. Thus, ℓ 2416. Inserting this into (32), and using Lemma 7, we get ≥ 1 2 log(2a) 2416 log α < = 4 log 2, − 1039 2a Xa≥1 a contradiction. Thus, t 1 so n > 1. We now put ≥ 1 := i : r m and = 1,...,t . I { i | } J { }\I We put
M = ri. iY∈I We also let j be minimal in . We split the sum appearing in (32) in two parts: J Sd = L1 + L2, Xd|n where L1 := Sd and L2 := Sd. d|n d|n r|d⇒Xr|2M ru|d forX some u∈J ′ To bound L1, we note that all divisors involved divide n , where
′ s λi n =2 ri . iY∈I Using Lemmas 7 and 8, we get log d L1 2 ≤ ′ d Xd|n log r n′ < 2 ′ ′ r 1 φ(n ) Xr|n − log r 2M (33) = 2 . r 1 φ(2M) rX|2M − We now bound L2. If = , then L2 = 0 and there is nothing to bound. So, assume that = . WeJ argue∅ as follows. Note that since s 1, by Lemma 2 (i), we have J 6 ∅ ≥
−1 P = P 1 Q 1 Q 1 Q s . n n n 2n ··· 2 n1 Let q be any odd prime factor of Qn1 . By reducing relation (ii) of Lemma 2 modulo q and using the fact that n and q are both odd, we get 2P 2 1 (mod q), therefore 1 n1 ≡ 2 = 1. Hence, z(q) q 1 for such primes q. Now let d be any divisor of n which q | − 1 is a multiple of rj . The number of them is τ(n1/rj ), where τ(u) is the number of divisors of the positive integer u. For each such d, there is a primitive prime factor q of Q Q 1 . Thus, r d q 1. This shows that d d | n j | | d − (34) ν (φ(P )) ν (φ(Q 1 )) τ(n /r ) τ(n )/2, rj n ≥ rj n ≥ 1 j ≥ 1 12 BERNADETTE FAYE AND FLORIAN LUCA where the last inequality follows from the fact that τ(n /r ) λ 1 1 j = j . τ(n1) λj +1 ≥ 2
Since rj does not divide m, it follows from (5) that (35) ν (P ) e . rj m ≤ rj Hence, (34), (35) and (1) imply that (36) τ(n ) 2e . 1 ≤ rj Invoking Lemma 5, we get
(rj + 1) log α (37) τ(n1) . ≤ log rj a Now every divisor d participating in L2 is of the form d = 2 d1, where 0 a s and d is a divisor of n divisible by r for some u . Thus, ≤ ≤ 1 1 u ∈ J
(38) L2 τ(n1) min S2ad1 := g(n1,s,r1). ≤ 0≤a≤s dX1|n1 ru|d1 for some u∈J In particular, d 3 and since the function x log x/x is decreasing for x 3, we 1 have that ≥ 7→ ≥ a log(2 rj ) (39) g(n1,s,r1) 2τ(n1) a . ≤ 2 rj 0≤Xa≤s s1 Putting also s1 := min s, 416 , we get, by Lemma 9, that 2 ℓ. Thus, inserting this as well as (33) and{ (39) all} into (32), we get |
1 log r 2M (40) ℓ log α < 2 + g(n ,s,r ). − 1039 r 1 φ(2M) 1 1 rX|2M − Since a log(2 rj ) 4 log 2 + 2 log rj (41) a < , 2 rj rj 0≤Xa≤s inequalities (41), (37) and (39) give us that 1 4 log 2 g(n ,s,r ) 2 1+ 2+ log α := g(r ). 1 1 ≤ r log r j j j The function g(x) is decreasing for x 3. Thus, g(rj ) g(3) < 10.64. For a positive integer N put ≥ ≤
1 log r N (42) f(N) := N log α 2 . − 1039 − r 1 φ(N) Xr|N − Then inequality (40) implies that both inequalities
f(ℓ)
s 1 1 Pn 1 2 M log α 39 ℓ log α 39 log log 1+ +L2, −10 ≤ −10 ≤ φ(Pn) ≤ s p 1 dX|2 M pX∈Pd − so
s 1 1 (45) L2 2 M log α 39 log 1+ . ≥ − 10 − s p 1 dX|2 M pX∈Pd − In the right–hand side above, M 1, 3, 5, 7 and s 1, 2, 3 . The values of the right–hand side above are in fact ∈ { } ∈ { } 1 h(u) := u log α log(P /φ(P )) − 1039 − u u for u =2sM 2, 4, 6, 8, 10, 12, 14, 20, 24, 28, 40, 56 . Computing we get: ∈{ } M h(u) H for M 1, 3, 5, 7 , s 1, 2, 3 , ≥ s,M φ(M) ∈{ } ∈{ } where
H1,1 > 1.069, H1,M > 2.81 for M > 1, H2,M > 2.426, H3,M > 5.8917. 14 BERNADETTE FAYE AND FLORIAN LUCA
We now exploit the relation M (46) H ω(n1) (rj + 1) log α 2 τ(n1) < rj , ≤ ≤ log rj where the last inequality holds because rj is large. Thus, log r (47) ω(n ) < j < 2 log r . 1 log 2 j Hence, n M 1 ω(n1) M 1 2 log rj 1 1+ < 1+ φ(n ) ≤ φ(M) r 1 φ(M) r 1 1 j − j − M 2 log r M 4 log r (48) < exp j < 1+ j , φ(M) r 1 φ(M) r 1 j − j − where we used the inequalities 1 + x 4 + 4 log log(8rj ) 4 log rj 1 M (49) L2 τ(n1) 1+ a . ≤ rj rj 1 φ(2 ) φ(M) − 0≤Xa≤s Inserting inequality (37) into (49) and using (46), we get 1 4 log r G (50) log r < 4 1+ 1+ j (1 + log log(8r ))(log α) s , j r r 1 j H j j − s,M where 1 G = . s φ(2a) 0≤Xa≤s PELL NUMBERS WHOSE EULER FUNCTION IS A PELL NUMBER 15 For s =2, 3, inequality (50) implies rj < 900, 000 and rj < 300, respectively. For s = 1 and M > 1, inequality (50) implies rj < 5000. When M = 1 and s = 1, we 12 get n = 2n1 and j = 1. Here, inequality (50) implies that r1 < 8 10 . This is too big, so we use the bound × 2 log d S < d d of Lemma 7 instead for the divisors d of participating in L2, which in this case are all the divisors of n larger than 2. We deduce that log d log d1 1.06 log r n1 4 log r1 4 log r1 1.06 < 4 < ω(n1) 1+ r 1 φ(n1) r1 1 r1 1 rX|n1 − − − 4 log r 4 log r < 1 (2 log r ) 1+ 1 , r 1 1 r 1 1 − 1 − 6 which gives r1 < 159. So, in all cases, rj < 10 . Here, we checked that er = 1 for all such r except r 13, 31 for which e =2. If e = 1, we then get τ(n /r ) 1, ∈{ } r rj 1 j ≤ so n = r . Thus, n 8 106, in contradiction with Lemma 9. Assume now that 1 j ≤ · rj 13, 31 . Say rj = 13. In this case, 79 and 599 divide Q13 which divides ∈ { } 2 Pn, therefore 13 (79 1)(599 1) φ(Pn) = Pm. Thus, if there is some other ′ | − − ′ | prime factor r of n1/13, then 13r n1, and Q13r′ has a primitive prime factor ′ | q 1 (mod 13r ). In particular, 13 q 1. Thus, ν13(φ(Pn)) 3, showing that 13≡3 P . Hence, 13 m, therefore 13| M−, a contradiction. A similar≥ contradiction | m | | is obtained if rj = 31 since Q31 has two primitive prime factors namely 424577 and 865087 so 31 M. | This finishes the proof. Acknowledgments B. F. thanks OWSD and Sida (Swedish International Development Cooperation Agency) for a scholarship during her Ph.D. studies at Wits. References [1] Y. F. Bilu, G. Hanrot and P. M. Voutier, “Existence of Primitive Divisors of Lucas and Lehmer Numbers. With an appendix by M. Mignotte”, J. Reine Angew. Math. 539 (2001), 75–122. [2] R. D. Carmichael, “On the numerical factors of the arithmetic forms αn ± βn”, Ann. Math. (2) 15 (1913), 30–70. [3] J. Cilleruelo and F. Luca, “Repunit Lehmer numbers”, Proc. Edinb. Math. Soc. (2) 54 (2011), no. 1, 55-65. [4] B. Faye, F. Luca and A. Tall, “On the equation φ(5m − 1) = 5n − 1”, Bull. Korean Math. Soc. 52 (2015), 513–524. [5] B. Faye and F. Luca, “Lucas numbers with the Lehmer property”, Preprint, 2015. 16 BERNADETTE FAYE AND FLORIAN LUCA [6] Robin Guy, “Estimation de la fonction de Tchebychef θ sur le k-ieme nombre premier et grandes valeurs de la fonction ω(n) nombre de diviseurs premiers de n”, Acta Arith. 42 (1983), no. 4, 367–389. [7] J. M. Grau Ribas and F. Luca,“Cullen numbers with the Lehmer property”, Proc. Amer. Math. Soc. 140 (2012), no. 1, 129–134: Corrigendum 141 (2013), no. 8, 2941–2943. [8] F. Luca, “Arithmetic functions of Fibonacci numbers”, Fibonacci Quart. 37 (1999), no. 3, 265–268. [9] F. Luca, “Multiply perfect numbers in Lucas sequences with odd parameters”, Publ. Math. Debrecen 58 (2001), no. 1-2, 121–155. [10] F. Luca, “Fibonacci numbers with the Lehmer property”, Bull. Pol. Acad. Sci. Math. 55 (2007), 7–15. [11] F. Luca and F. Nicolae,“φ(Fm)= Fn”, Integers 9 (2009), A30. [12] W. McDaniel, “On Fibonacci and Pell Numbers of the Form kx2”, Fibonacci Quart. 40, (2002), 41–42. [13] H. L. Montgomery and R. C. Vaughan, “The large sieve”, Mathematika 20 (1973), 119–134. [14] A. Peth˝o, “The Pell sequence contains only trivial perfect powers”, in Sets, graphs and numbers (Budapest, 1991), Colloq. Math. Soc. Janos Bolyai 60, North-Holland, Amsterdam, (1992), 561–568. [15] J. B. Rosser, L. Schoenfeld, “Approximate formulas for some functions of prime numbers“, Illinois J. Math. 6 (1962), 64–94. Ecole Doctorale de Mathematiques et d’Informatique Universite´ Cheikh Anta Diop de Dakar BP 5005, Dakar Fann, Senegal and School of Mathematics University of the Witwatersrand Private Bag X3, Wits 2050, South Africa E-mail address: [email protected] School of Mathematics, University of the Witwatersrand Private Bag X3, Wits 2050, South Africa E-mail address: [email protected]