On Strings of Consecutive Integers with a Distinct Number of Prime Factors

Let ω(n) be the number of distinct prime factors of n. For any positive integer k let n = nk be the smallest positive integer such that ω(n + 1), . . . , ω(n + k) are mutually distinct. In this paper, we give upper and lower bounds for nk. We study the same quantity when ω(n) is replaced by Ω(n), the total number of prime factors of n counted with repetitions. Let ω(n) and Ω(n) denote respectively the number of distinct prime factors of n and the total number of prime factors of n counted with repetitions. For any positive integer k let n = nk be the smallest positive integer n such that ω(n+1), . . . , ω(n+k) are mutually distinct. We also let m = mk be the smallest positive integer m such that Ω(m + 1), . . . ,Ω(m + k) are mutually distinct. Using a computer, we easily obtain that n2 = 4, n3 = 27, n4 = 416, n5 = 14321, n6 = 461889, n7 = 46908263 and n8 = 7362724274, and also that m2 = 2, m3 = 5, m4 = 14, m5 = 59, m6 = 725, m7 = 6317, m8 = 189374, m9 = 755967 and m10 = 683441870. In this paper, we give upper and lower bounds for nk and mk. Let pi be the i-th prime number. Let n = nk. Since the set {ω(n+ j) : j = 1, . . . , k} consists of k nonnegative integers, it follows that one of n+ j for j = 1, . . . , k must have at least k distinct prime factors. Thus, n+ k ≥ k ∏ i=1 pi = exp((1 + o(1))pk) = exp((1 + o(1))k log k) as k → ∞ by the Prime Number Theorem; therefore nk ≥ exp((1 + o(1))k log k) as k → ∞. Similarly, letting m = mk, we get that Ω(m+ i) ≥ k for some i ∈ {1, . . . , k}. Thus, m+ k ≥ 2, giving mk ≥ exp((log 2 + o(1))k) as k → ∞. We start by improving these trivial estimates as follows. Theorem 1. The inequality nk ≥ exp((2 + o(1))k log k) holds as k → ∞. Furthermore, the inequality mk ≥ exp((1/2 + o(1))k log k) holds as k → ∞. Received by the editors May 16, 2008, and, in revised form, July 3, 2008. 2000 Mathematics Subject Classification. Primary 11A25, 11N64. c ©2008 American Mathematical Society 1585 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1586 JEAN-MARIE DE KONINCK, JOHN B. FRIEDLANDER, AND FLORIAN LUCA The problem of finding lower and upper bounds for nk and mk was raised in the recent book [1] by the first author. We remark that, after writing this paper, we noticed that the first of these bounds is essentially equivalent to one due to Erdős [2]. We were somewhat surprised that we could not find any other work on these problems. Proof. We start with the first inequality. Assume that ω(n + 1), . . . , ω(n + k) are mutually distinct. Let ε ∈ (0, 1) be arbitrarily small but fixed. Put s = k1−ε . Let i1, . . . , is be s distinct integers in {1, . . . , k} such that ω(n + ij) ≥ k − j for j = 1, . . . , s. Let Aij be the set of prime factors of n + ij . Note that if j = and p ∈ Aij ∩Ai , then p | (n+ ij)− (n+ i ) = (ij − i ) and 1 ≤ |ij − i | ≤ k− 1. Since ω(m) logm/ log logm holds for all positive integers, we get that # ( Aij ∩ Ai ) < c1 log k log log k holds for all j = with some absolute constant c1. By the Principle of Inclusion and Exclusion,