Consider the following congruence: For $p\geq 5$ prime and every $n,\nu\in\mathbb{N}$ we have \begin{align*} 0\equiv\sum_{k=1\atop p\nmid k}^{pn1}\frac1k \binom{pn(\nu+1)k1}{pn\nu1} \mod p^{2(\operatorname{ord}_p(n)+1)}\mathbb{Z}_p. \end{align*} My questions: Is this a known identity? Can it be obtained by a more general statement?
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1$\begingroup$ Somewhat similar congruence and related references are given in my paper arxiv.org/abs/1602.02632 $\endgroup$ – Max Alekseyev Mar 12 at 22:45