MathNet 2020 — Math Olympiad Vault — Part 01 Question 45

Let $a_1 = 2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_n$ that has more positive divisors than $a_n$. Prove that $2a_{n+1} = 3a_n$ only for finitely many indices $n$.

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