ON A DENSITY PROBLEM OF ERDÖS

For a positive integer n, let P(n) denotes the largest prime divisor of n and define the set: (x)= = {n≤ x :n does not divide P(n)!}. Paul Erdös has proposed that |S| = o(x) as x→∞, where |S| is the number of n∈ S. This was proved by Ilias Kastanas. In this paper we will show the stronger result that |S| =O(xe−1/4√logx).

Introduction. For a positive integer n, let P (n) denote the largest prime divisor of n and define the set (x) = = n ≤ x : n does not divide P (n)!}. (1) Paul Erdös [1] proposed that |S| = o(x) as x → ∞, where |S| is the number of n ∈ S. A solution [3] was provided by Ilias Kastanas. There was also [3] a claim of proving that |S(x)| = O(x/log x). In this paper, we show the stronger result.
Theorem. For some constant a > 0, we have In fact, a = 1/4 suffices. Lemma 1. Let ν(n) be the number of distinct prime divisors of n. Define Then Proof. It is well known [2] that if d(m) is the number of divisors of m, then and the lemma follows.
will be chosen later. Define Then Proof. Since n ∈ S 2 if and only if n = tp 2 for some C < p ≤ √ x and some t ≤ x/p 2 , The first big O in (8) follows since where T = 2 log C. Then Proof.
, then, for any n ∈ S , we have Proof. Let n ∈ S . Then n ∈ S and so n does not divide P (n)!. There exists a prime p 0 dividing n such that where ν p (m) denotes the largest integer t such that p t divides m. Since ν p 0 (P (n)!) ≥ 1, (13) implies that ν p 0 (n) ≥ 2. Since n ∈ S 2 , it follows that p 0 ≤ C. Also n ∈ S 3 , so that T ≥ ν p 0 (n). Hence, which implies (12). Note that the third inequality in (14) is true because ν p (m!) ≥ m/2p for any m and any p | m. This is because and Proof of the Theorem. For n ∈ S , we have n ∈ S 1 ∪ S 2 ∪ S 3 . Thus, ν p (n) ≤ 4k log log x. Also, if p α is any prime power dividing n, then one of the following two possibilities must occur: (a) p ≤ C and α ≤ T , (b) p > C and α = 0 or 1. Case (a) generates at most C(T + 1) ≤ 2CT prime powers. For Case (b), the number of prime powers p α with p > C and α ≤ 1 is at most P (n). By Lemma 4, this is at most 2CT . Hence, the number of possible prime powers p α that divide an n ∈ S is at most 4CT . But such an n can be the product of at most 4k log log x distinct prime powers. Therefore, Hence, (18) and (19) yield and (2) follows with a = 1/4.

Remark.
If π(x) is the number of prime integers that are less than or equal to x, an early version of the prime numbers theorem asserts that for some constant a. Although the big O terms in (19) and (2) are similar, there is no apparent relationship between the PNT and (2).