CN-TST 2021 Problem 16

Proof that $\sum_{m=1}^n5^{\omega (m)} \le \sum_{k=1}^n\lfloor \frac{n}{k} \rfloor \tau (k)^2 \le \sum_{m=1}^n5^{\Omega (m)} .$

Dataset text AI-MO AoPS. Quoted with source.

Solutions

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