Abstract

We prove several congruences modulo a power of a prime, such as

∑_{0<k_{1}<···<k_{n}<p} (\frac{p − k_{n}}{3})\frac{(−1)^{k_{n}}}{k_{1} ⋯ k_{n}} ≡ \begin{cases}−\frac{2^{n+1} + 2}{6^{n+1}} p B_{p−n−1} (1/3) (mod p^{2}), & if n is odd, \\ −\frac{2^{n+1} + 4}{n6^{n}} B_{p−n} (1/3) (mod p), & if n is even, \end{cases}

where n is a positive integer and p is a prime such that p > max(n + 1, 3).