Published 04-03-2025
Keywords
- harmonic numbers,
- Euler number,
- congruences
Abstract
In this short note, our primary purpose is to prove the congruence
\[\sum_{k=1}^{\frac{p-1}{2}} \frac{(-1)^k}{k} \sum_{j=\lfloor\frac{k}{2}\rfloor+1}^{k} \frac{1}{2j-1} \equiv 0 \mod p\]
Along the way, a number of auxiliary results of independent interest are found.
