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sum of max prime divisors of n such that P^k <=n

Source: 2022 Argentina OMA Finals L3 p1

March 25, 2024
number theory

Problem Statement

For every positive integer nn, P(n)P(n) is defined as follows: For each prime divisor pp of nn is considered the largest integer kk such that pk≤np^k\le n and all the pkp^k are added. For example, for n=100=22⋅52n=100=2^2 \cdot 5^2, as 26<100<272^6<100<2^7 and 52<100<535^2<100<5^3, it turns out that P(100)=26+52=89P(100)=2^6+5^2=89 Prove that there are infinitely many positive integers nn such that P(n)>nP(n)>n..
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