MathDB

For every natural number

n

we define

f(n)

by the following rule:

f(1) = 1

and for

n>1

then

f(n) = 1 + a_1 \cdot p_1 + \ldots + a_k \cdot p_k

, where

n = p_1^{a_1} \cdots p_k^{a_k}

is the canonical prime factorisation of

n

(

p_1, \ldots, p_k

are distinct primes and

a_1, \ldots, a_k

are positive integers). For every positive integer

s

, let

f_s(n) = f(f(\ldots f(n))\ldots)

, where on the right hand side there are exactly

s

symbols

f

. Show that for every given natural number

a

, there is a natural number

s_0

such that for all

s > s_0

, the sum

f_s(a) + f_{s-1}(a)

does not depend on

s

Problem Statement