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.
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.