Abstract

We prove that $$ \sum_{p \leq x} \frac{1}{\tau(p-1)} \asymp \frac{x}{(\log x)^{3/2}} \quadand\quad \sum_{n \leq x} \frac{1}{\tau(n^2+1)} \asymp \frac{x}{(\log x)^{1/2}}, $$ where $\tau(n)=\sum_{d\mid n}1$ is the number of divisors of $n$, and the first sum is taken over prime numbers. Bibliography: 14 titles.

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