THE GREATEST PRIME DIVISOR OF A PRODUCT OF CONSECUTIVE INTEGERS SHANTA LAISHRAM AND T. N. SHOREY 1. Introduction Let k ≥ 2 and n ≥ 1 be integers. We denote by Δ(n, k)= n(n + 1) ··· (n + k − 1). For an integer ν> 1, we denote by ω(ν ) and P (ν ) the number of distinct prime divisors of ν and the greatest prime factor of ν , respectively, and we put ω(1) = 0, P (1) = 1. A well known theorem of Sylvester [7] states that P (Δ(n, k)) >k if n > k. (1) We observe that P (Δ(1,k)) ≤ k and therefore, the assumption n>k in (1) cannot be removed. For n>k, Moser [5] sharpened (1) to P (Δ(n, k)) > 11 10 k and Hanson [3] to P (Δ(n, k)) > 1.5k unless (n, k) = (3, 2), (8, 2), (6, 5). Further Faulkner [2] proved that P (Δ(n, k)) > 2k if n is greater than or equal to the least prime exceeding 2k and (n, k) = (8, 2), (8, 3). In this paper, we sharpen the results of Hanson and Faulkner. We shall not use these results in the proofs of our improvements. We prove Theorem 1. We have (a) P (Δ(n, k)) > 2k for n> max(k + 13, 279 262 k). (2) (b) P (Δ(n, k)) > 1.97k for n>k + 13. (3) We observe that 1.97 in (3) cannot be replaced by 2 since there are arbitrary long chains of consecutive composite positive integers. The same reason implies that Theorem 1 (a) is not valid under the assumption n>k + 13. Further the assumption n> 279 262 k in Theorem 1 (a) is necessary since P (Δ(279, 262)) ≤ 2 × 262. 2000 Mathematics Subject Classification: Primary 11A41, 11N05, 11N13. Keywords: Arithmetic Progressions, Primes. 1