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Problem 4, BMO 2020

Source: Problem 4, BMO 2020

November 1, 2020
number theoryBMO

Problem Statement

Let a1=2a_1=2 and, for every positive integer nn, let an+1a_{n+1} be the smallest integer strictly greater than ana_n that has more positive divisors than ana_n. Prove that 2an+1=3an2a_{n+1}=3a_n only for finitely many indicies nn.
Proposed by Ilija Jovčevski, North Macedonia
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