Supercongruences motivated by e

preprint2015arXivOpen access

Abstract

In this paper we establish some new supercongruences motivated by the well-known fact lim⁡n→∞(1+1/n)n=e\lim_{n\to\infty}(1+1/n)^n=e. Let p>3p>3 be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{-1/(p+1)}k^{p+1}\equiv 0\ \pmod{p^5}\ \ \ \mbox{and}\ \ \ \sum_{k=0}^{p-1}\binom{1/(p-1)}k^{p-1}\equiv \frac{2}{3}p^4B_{p-3}\ \pmod{p^5},$$ where B0,B1,B2,…B_0,B_1,B_2,\ldots are Bernoulli numbers. We also show that for any a∈Za\in\mathbb Z with p∤ap\nmid a we have $$\sum_{k=1}^{p-1}\frac1k\left(1+\frac ak\right)^k\equiv -1\pmod{p}\ \ \ \mbox{and}\ \ \ \sum_{k=1}^{p-1}\frac1{k^2}\left(1+\frac ak\right)^k\equiv 1+\frac 1{2a}\pmod{p}.$$

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