On the denominators of harmonic numbers ∗ Bing-Ling Wu and Yong-Gao Chen† School of Mathematical Sciences and Institute of Mathematics, Nanjing Normal University, Nanjing 210023, P. R. China Abstract. Let Hn be the n-th harmonic number and let vn be its denomi- nator. It is well known that vn is even for every integer n ≥ 2. In this paper, we study the properties of vn. One of our results is: the set of positive integers 1/4 n such that vn is divisible by the least common multiple of 1, 2, ··· , ⌊n ⌋ has density one. In particular, for any positive integer m, the set of positive integers n such that vn is divisible by m has density one. 2010 Mathematics Subject Classification: 11B75, 11B83 Keywords and phrases: Harmonic numbers; p-adic valuation; Asymptotic density 1 Introduction For any positive integer n, let arXiv:1711.00184v1 [math.NT] 1 Nov 2017 1 1 1 un Hn =1+ + + ··· + = , (un, vn)=1, vn > 0. 2 3 n vn The number Hn is called n-th harmonic number. In 1991, Eswarathasan and Levine [2] introduced Ip and Jp. For any prime number p, let Jp be the set of ∗This work was supported by the National Natural Science Foundation of China (No. 11771211) and a project funded by the Priority Academic Program Development of Jiangsu Higher Education Institutions. †Corresponding author, E-mail: [email protected] (B.-L. Wu), [email protected](Y.-G. Chen) 1 positive integers n such that p | un and let Ip be the set of positive integers n such that p ∤ vn. Here Ip and Jp are slightly different from those in [2]. In [2], Eswarathasan and Levine considered 0 ∈ Ip and 0 ∈ Jp. It is clear that Jp ⊆ Ip. In 1991, Eswarathasan and Levine [2] conjectured that Jp is finite for any prime number p. In 1994, Boyd [1] confirmed that Jp is finite for p ≤ 547 except 83, 127, 397. For any set S of positive integers, let S(x) = |S ∩ [1, x]|. In 2016, Sanna [3] proved that 2 0.765 Jp(x) ≤ 129p 3 x . Recently, Wu and Chen [4] proved that 2 + 1 p Jp(x) ≤ 3x 3 25 log . (1.1) For any positive integer m, let Im be the set of positive integers n such that m ∤ vn. In this paper, the following results are proved. Theorem 1.1. The set of positive integers n such that vn is divisible by the least common multiple of 1, 2, ··· , ⌊n1/4⌋ has density one. Theorem 1.2. For any positive integer m and any positive real number x, we have 1 2 + 1 q Im(x) ≤ 4m 3 x 3 25 log m , where qm is the least prime factor of m. From Theorem 1.1 or Theorem 1.2, we immediately have the following corol- lary. Corollary 1.3. For any positive integer m, the set of positive integers n such that m | vn has density one. 2 Proofs We always use p to denote a prime. Firstly, we give the following two lemmas. Lemma 2.1. For any prime p and any positive integer k, we have k k Ipk = {p n1 + r : n1 ∈ Jp ∪{0}, 0 ≤ r ≤ p − 1}\{0}. 2 Proof. For any integer a, let νp(a) be the p-adic valuation of a. For any rational a number α = b , let νp(α) = νp(a) − νp(b). It is clear that n ∈ Ipk if and only if νp(Hn) > −k. k If n<p , then νp(Hn) ≥ −νp([1, 2, ··· , n]) > −k. So n ∈ Ipk . In the following, we assume that n ≥ pk. Let k k n = p n1 + r, 0 ≤ r ≤ p − 1, n1,r ∈ Z. Then n1 ≥ 1. Write n 1 1 b u pbv + au H = + H = + n1 = n1 n1 , (2.1) n m pk n1 pk−1a pkv pkav =1 k n1 n1 m X,p ∤m where p ∤ a and (un1 , vn1 ) = 1. k If n1 ∈ Jp, then p | un1 and p ∤ vn1 . Thus p | aun1 + pbvn1 and νp(p avn1 )= k. By (2.1), νp(Hn) > −k. So n ∈ Ipk . If n1 ∈/ Jp, then p ∤ un1 . Thus p ∤ aun1 + pbvn1 . It follows from (2.1) that νp(Hn) ≤ −k. So n∈ / Ipk . Now we have proved that n ∈ Ipk if and only if n1 ∈ Jp ∪{0}. This completes the proof of Lemma 2.1. Lemma 2.2. For any prime power pk and any positive number x, we have 1 − 1 2 + 1 k 3 25 log p 3 25 log p Ipk (x) ≤ 4(p ) x . Proof. If x ≤ pk, then 1 − 1 2 + 1 1 − 1 2 + 1 3 25 log p 3 25 log p k 3 25 log p 3 25 log p Ipk (x) ≤ x< 4x x ≤ 4(p ) x . Now we assume that x>pk. By Lemma 2.1 and (1.1), we have k k Ipk (x) = |{p n1 + r ≤ x : n1 ∈ Jp ∪{0}, 0 ≤ r ≤ p − 1}| − 1 x 1 − 1 2 + 1 ≤ pk J ( )+1 ≤ 4(pk) 3 25 log p x 3 25 log p .. p pk This completes the proof of Lemma 2.2. θ Proof of Theorem 1.1. Let mn be the least common multiple of 1, 2, ··· , ⌊n ⌋ and let ⌊nθ⌋ denote the greatest integer not exceeding the real number nθ, where 0 <θ< 1 which will be given later. Let T = {n : mn ∤ vn}. For any prime p and 3 θ αp θ any positive number x with p ≤ x , let αp be the integer such that p ≤ x < pαp+1. By the definition of mn and T , T (x) ≤ Ipαp (x). θ pX≤x Thus, by Lemma 2.2, we have 1 2 + 1 αp p Ipαp (x) ≤ 4 (p ) 3 x 3 25 log := I1 + I2, θ θ pX≤x pX≤x where 1 2 1 1 2 1 αp + αp + I1 =4 (p ) 3 x 3 25 log p , I2 =4 (p ) 3 x 3 25 log p δ θ δ x <pX≤x pX≤x and δ is a positive constant less than θ which will be given later. If p > xδ, then 1 log x log x 1 x 25 log p = e 25 log p ≤ e 25δ log x = e 25δ . It follows from pαp ≤ xθ that 1 2 1 1 θ 2 4θ 2 αp + + 1 + I1 =4 (p ) 3 x 3 25 log p ≤ 4e 25δ x 3 3 ≪ x 3 3 . δ θ δ θ log x x <pX≤x x <pX≤x α θ For I2, by p p ≤ x we have 1 2 1 αp + I2 = 4 (p ) 3 x 3 25 log p δ pX≤x θ + 2 + 1 ≤ 4 x 3 3 25 log 2 δ pX≤x 1 δ+ θ + 2 + 1 ≪ x 3 3 25 log 2 . log x 1 We choose θ = 4 and δ =0.1. Then x 0.91 I1 ≪ , I2 ≪ x . log x Therefore, x T (x) ≤ Ipαp (x)= I1 + I2 ≪ . θ log x pX≤x It follows that the set of positive integers n such that vn is divisible by the least common multiple of 1, 2, ··· , ⌊n1/4⌋ has density one. This completes the proof of Theorem 1.1. 4 Proof of Theorem 1.2. We use induction on m to prove Theorem 1.2. By Lemma 2.2, Theorem 1.2 is true for m = 2. Suppose that Theorem 1.2 is true for all integers less than m (m> 2). If x ≤ m, then 1 2 + 1 1 2 + 1 q q Im(x) ≤ x< 4x 3 x 3 25 log m ≤ 4m 3 x 3 25 log m . In the following, we always assume that x > m. If m is a prime power, then, by Lemma 2.2, Theorem 1.2 is true. Now we assume that m is not a prime power. Let pα be the least prime power divisor α α of m such that m = p m1 with p ∤ m1. Then m1 > p . It is clear that Im = α α α Im1 (Ip \Im1 ). By Lemma 2.1 and the definition of p , {1, 2, ··· ,p −1} ⊆ Im1 . HenceS α α α Im(x)= Im1 (x)+(Ip \ Im1 )(x) ≤ Im1 (x)+ Ip (x) − (p − 1). By the inductive hypothesis, we have 1 2 + 1 1 2 + 1 3 3 25 log qm1 3 3 25 log qm Im1 (x) ≤ 4m1 x ≤ 4m1 x . It follows that 1 2 + 1 3 3 25 log q α Im(x) ≤ 4m1 x m + Ipα (x) − (p − 1). (2.2) We divide into the following three cases: α α Case 1: p ≥ 8. Then m1 >p ≥ 8. By Lemma 2.2, we have 1 2 + 1 α q Ipα (x) ≤ 4(p ) 3 x 3 25 log m . It follows from (2.2) that 1 2 + 1 1 2 + 1 3 q α q Im(x) ≤ 4m1 x 3 25 log m + 4(p ) 3 x 3 25 log m 1 1 1 2 + 1 3 3 25 log qm = 4 1 + 1 m x α 3 3 (p ) m1 ! 1 2 + 1 ≤ 4m 3 x 3 25 log qm . Case 2: pα < 8, p = 2. Then pα = 2 or 4 and x > m ≥ 2 × 3 = 6. By Lemma 2.1 and J2 = ∅, we have I4 = {1, 2, 3} and I2 = {1}. It is clear that α Ipα (x) − (p − 1) = 0. It follows from (2.2) that 1 2 + 1 1 2 + 1 3 q q Im(x) ≤ 4m1 x 3 25 log m < 4m 3 x 3 25 log m .