How many ordered factorizations may n have?

Martin Klazar

KAM, MFF UK

Abstract

If n>0 is an integer, it is easy to show that it may be expressed in exactly 2^{n-1} ways as an ordered sum (i.e., the order of summands maters) of positive integers, n = a_1 + a_2 + ... + a_k. We replace sums by products and denote m(n) the number of ordered factorizations n = a_1 * a_2 * ... * a_k of n into integers a_i>1. For example, m(12)=8 because of the factorizations 12, 2*6, 6*2, 3*4, 4*3, 2*2*3, 2*3*2, 3*2*2. The function m(n) stronly depends on prime factorization of n; for example, m(n)=1 whenever n is a prime number.

In my talk I will discuss the problem what is the maximum value of m(n) in terms of n. We will see that the maximum is, roughly, n^{1.728...} where s=1.728... is the (positive real) solution to zeta(s) = 2 (zeta(s) = 1^{-s} + 2^{-s} + 3^{-s} + ...). I will explain more precise estimates of the maximum obtained jointly with F. Luca, as well as the recent ultimate improvement