Abstract

has d(k!) d((k I)!) solutions, where d(n) denotes the number of divisors of n. This follows from {x: S(x) = k} = {x: xlk!, x t (k I)!}. Thus, equation (1) always has at least a solution, if d(k!) > d((k I)!) for k ~ 2. In what follows, we shall prove this inequality, and in fact we will consider the arithmetical functions y, <7, d, w, 0 at factorials. Here y( n) = Euler's arithmetical function, <7( n) = sum of divisors of n, w( n) = number of distinct prime factors of n, O( n) = number of total divisors of n. As it is well known, r we have y( 1) = d(1) = 1. while w(l) = O( 1) = 0, and for 1 < IT pf; (ai ~ 1, Pi distinct i=l primes) one has y( n) = n IT (1 -~) , i=l p, r a,+l 0( n) = IT Pi 1 , P -1 i=l ' w(n) = r,

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