On a density problem of Erdös

Safwan Akbik

doi:10.1155/s0161171299226555

Abstract

For a positive integer n, let P(n) denotes the largest prime divisor of n and define the set: 𝒮(x) = 𝒮 = {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 .

Highlights

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
The first big O in (8) follows since p>C 1/p2 ≤
The first inequality in (11) is valid because α≥T 1/pα ≤ 1/pT + 1/pT +1 + · · · ≤ 2/pT and the second is valid because pT = 2T (p/2)T ≥ 2T (p/2)2 for T ≥ 2

Summary

Introduction

For a positive integer n, let P (n) denote the largest prime divisor of n and define the set Let ν(n) be the number of distinct prime divisors of n. It is well known [2] that if d(m) is the number of divisors of m, m≤x d(m) = O(x log x).

Results

Conclusion