Let $p$ be any prime and let $a$ and $n$ be positive integers with $p\nmid n$. We show that $$\sum_{ k=1}^{ p^a-1}\frac{ B_k}{ (-n)^k}\equiv a(-1)^{ n-1}D_{ n-1}\pmod { p},$$ where $B_0,B_1,\ldots$ are the Bell numbers and $D_0,D_1,\ldots$ are the derangement numbers. This extends a result of Sun and Zagier published in 2011. Furthermore, we prove that $$(-x)^n\sum_{ k=1}^{ p^a-1}\frac{ B_k(x)}{ (-n)^k}\equiv -\sum_{ r=1}^ax^{ p^r}\sum_{ k=0}^{ n-1}\frac{ (n-1)!}{ k!}(-x)^k\pmod{ p\mathbb Z_p[x]},$$ where $B_k(x)=\sum_{ l=0}^kS(k,l)x^l$ is the Bell polynomial of degree $k$ with $S(k,l)\ (0\le l\le k)$ the Stirling numbers of the second kind, and $\mathbb Z_p$ is the ring of all $p$-adic integers.
Categories: math.CO