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Almost perfect numbers

Source: European Mathematical Cup, 2015, Junior, P3

December 30, 2016
number theorydivisor

Problem Statement

Let d(n)d(n) denote the number of positive divisors of nn. For positive integer nn we define f(n)f(n) as f(n)=d(k1)+d(k2)++d(km),f(n) = d\left(k_1\right) + d\left(k_2\right)+ \cdots + d\left(k_m\right), where 1=k1<k2<<km=n1 = k_1 < k_2 < \cdots < k_m = n are all divisors of the number nn. We call an integer n>1n > 1 almost perfect if f(n)=nf(n) = n. Find all almost perfect numbers.
Paulius Ašvydis