International Mathematical Forum, Vol. 12, 2017, no. 13, 643 - 649 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/imf.2017.7543 Logarithm of the Exponents in the Prime Factorization of the Factorial Rafael Jakimczuk Divisi´onMatem´atica,Universidad Nacional de Luj´an Buenos Aires, Argentina Copyright c 2017 Rafael Jakimczuk. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduc- tion in any medium, provided the original work is properly cited. Abstract P In this note we study the sum 2≤p≤n log E(p) where E(p) is the exponent of the prime p in the prime factorization of n!, the sum P 2≤p≤n HE(p), where Hk denotes the k-th harmonic number and an- other sums. We also consider the generalized harmonic numbers Hk;m of order m and obtain a strong connection between these sums and the Riemann zeta function. Mathematics Subject Classification: 11A99, 11B99 Keywords: Factorial, prime factorization, Exponents, Logarithm 1 Introduction Let us consider the prime factorization of a positive integer a s1 s2 sk a = p1 p2 ··· pk ; where p1; p2; : : : ; pk are the different primes in the prime factorization and s1; s2; : : : ; sk are the exponents. The total number of prime factors in the prime factorization is denoted Ω(a) (see [1, chapter XXII]), that is, Ω(a) = s1 + s2 + ··· + sk: 644 Rafael Jakimczuk Let us consider the factorial n!, the exponent of a prime p in its prime factor- ization will be denoted E(p). Then we can write the prime factorization of n! in the form n! = Y pE(p) 2≤p≤n since, clearly, the primes that appear in the prime factorization of n! are the primes not exceeding n. It is well-known the following asymptotic formula (see [1, chapter XXII]) Ω(n!) = X E(p) = n log log n + An + o(n); (1) 2≤p≤n P where A is a constant. Note that Ω(n!) = 2≤a≤n Ω(a). In this note we obtain an asymptotic formula for the sequence (compare with (1)) L(n!) = X log E(p): 2≤p≤n Pk 1 Let us consider the k-th harmonic number Hk, namely Hk = i=1 i . We obtain an asymptotic formula for the sequence (compare with (1)) X M(n!) = HE(p): 2≤p≤n We also consider the k-th generalized harmonic number of order m ≥ 2, Hk;m, Pk 1 namely Hk;m = i=1 im and we obtain an asymptotic formula for the sequence (compare with (1)) X Mm(n!) = HE(p);m: 2≤p≤n where the Riemann zeta function appear. Finally, we consider the sum 1 X : 2≤p≤n (E(p) − 1)! where the e number appear. 2 Main Results Theorem 2.1 We have the asymptotic formula n n ! L(n!) = X log E(p) = C + o = Cπ(n) + o(π(n)); (2) 2≤p≤n log n log n Logarithm of the exponents of the primes in the factorial 645 where the constant C is 1 log j C = X : (3) j=2 j(j + 1) Proof. Note that, if we put 1 1 ! log j A = log j − = ; (4) j j j + 1 j(j + 1) the series of positive terms Aj converges, that is 1 X Aj = C: (5) j=1 We have (prime number theorem) x x ! π(x) = X 1 = + o : (6) 2≤p≤x log x log x Another more precise well-known formula is ! X x x π(x) = 1 = + f1(x) 2 ; (7) 2≤p≤x log x log x where jf1(x)j < M. We also need the following well-known formula X x #(x) = log p = x + f2(x) ; (8) 2≤p≤x log x where jf2(x)j < M. On the other hand, the exponent E(p) of the prime p in the prime factor- ization of n! is (Legendre's theorem) 1 $ % X n E(p) = j ; (9) j=1 p where b:c denotes the integer part function. If p satisfies the inequality (where j is a fixed positive integer) n n < p ≤ ; (10) j + 1 j and the inequality p p > n; (11) 646 Rafael Jakimczuk then we have 1 $ % $ % X n n E(p) = j = = j: (12) j=1 p p Note that if n is sufficiently large inequalities (10) and (11) are fulfilled since p n n < j+1 . Let > 0, we choose the fixed positive integer s such that the following inequalities hold 1 X 0 ≤ Aj ≤ , (13) j=s log s 0 ≤ ≤ , (14) s M ≤ . (15) s Now, we have (see (4), (5), (6), (10) and (12)) 0 1 s−1 X X X B X C L(n!) = log E(p) = log E(p) + @ log jA 2≤p≤n n j=1 n n 2≤p≤ s j+1 <p≤ j s−1 ! X n X n X n = log E(p) + Aj + o = log E(p) + C n log n j=1 log n n log n 2≤p≤ s 2≤p≤ s 1 n X n − Aj + o(1) : (16) log n j=s log n Besides, we have (see (7), (8) and (12)) 1 $ %! 1 ! X X X n X X n n log E(p) = log k ≤ log k = log nπ n n k=1 p n k=1 p s 2≤p≤ s 2≤p≤ s 2≤p≤ s n p ! − X log(p − 1) = log nπ − X log p + X log n s n n p − 1 2≤p≤ s 2≤p≤ s 2≤p≤ s n n n ≤ log nπ − # + π ; (17) s s s p since 1 < p−1 ≤ 2 < e. Let us consider the function log n f3(n) = n : (18) log s Logarithm of the exponents of the primes in the factorial 647 Note that limn!1f3(n) = 1. Therefore (see (8) and (18)) n n n 1 n # = + f f (n): (19) s s 2 s s log n 3 On the other hand, we have (see (7) and (18)) n n n n 1 n 1 π = + f1 2 = s s log n s n s log n 1 − log s s s log s log n n 1 n 1 n log s ! + f (f (n))2 = 1 + f (n) 1 s s log2 n 3 s log n log n 4 n 1 n 1 n log s n + f (f (n))2 = + f (n) 1 s s log2 n 3 s log n s log2 n 4 n 1 n 1 n + f (f (n))2 = f (n); (20) 1 s s log2 n 3 s log n 5 where limn!1f4(n) = 1, limn!1f5(n) = 1 and we have used the formula 1 1−x = 1 + x(1 + o(1)) (x ! 0). Substituting (19) and (20) into (17) we obtain (see (14) and (15)) X n n n log s n log E(p) ≤ log nπ − # + π = f4(n) n s s s s log n 2≤p≤ s n 1 n n 1 n 1 n + f (f (n))2 − f f (n) + f (n) 1 s 3 s log n 2 s 3 s log n 5 s log n log s n n 2 1 n n 1 n ≤ jf (n)j + f (jf (n)j) + f jf (n)j 4 s log n 1 s 3 s log n 2 s 3 s log n 1 n log s M M 1! n n + jf (n)j ≤ 2 + 4 + 2 + 2 ≤ 10 (21) 5 s log n s s s s log n log n Note that there exists n0 such that if n ≥ n0 then jfi(n)j ≤ 2 (i = 3; 4; 5), since fi(n) ! 1 (i = 3; 4; 5). Finally, equations (16), (13) and (21) give P n log E(p) 1 L(n!) 2≤p≤ s X − C ≤ + A + jo(1)j ≤ 12 (n ≥ n ) (22) n n j 0 log n log n j=s Note that there exists n0 such that if n ≥ n0 the o(1) (see equation (22)) satisfies jo(1)j ≤ . Therefore (2) is proved, since can be arbitrarily small. The theorem is proved. 648 Rafael Jakimczuk Theorem 2.2 We have the asymptotic formula 2 ! 2 X π n n π M(n!) = HE(p) = + o = π(n) + o(π(n)) 2≤p≤n 6 log n log n 6 Proof. The proof is the same as the proof of Theorem 2.1. Note that 1 H 1 1 1 1 1 1 1 1 X j = X + X + X + ··· j=1 j(j + 1) j=1 j(j + 1) 2 j=2 j(j + 1) 3 j=3 j(j + 1) 1 1 π2 = 1 + + + ··· = ζ(2) = 22 32 6 since 1 1 1 1 1 ! 1 X = X − = j=s j(j + 1) j=s j j + 1 s Besides we have the inequality 1 1 1 Z k 1 + + ··· + < dx = log k (k ≥ 2) 2 3 k 1 x and consequently the inequality Hk ≤ 1 + log k (k ≥ 1) Therefore (see (17)) X X n n n HE(p) ≤ (1 + log E(p)) ≤ log nπ − # + 2π n n s s s 2≤p≤ s 2≤p≤ s The theorem is proved. We have the following generalization of Theorem 2.2 where the generalized harmonic number replaces the harmonic number. Theorem 2.3 We have the asymptotic formula ! X n n Mm(n!) = HE(p);m = ζ(m + 1) + o = ζ(m + 1)π(n) + o(π(n)) 2≤p≤n log n log n Proof.