English

Supercongruences motivated by e

Number Theory 2015-02-27 v8 Combinatorics

Abstract

In this paper we establish some new supercongruences motivated by the well-known fact limn(1+1/n)n=e\lim_{n\to\infty}(1+1/n)^n=e. Let p>3p>3 be a prime. We prove that k=0p1(1/(p+1)k)p+10 (modp5)   \mboxand   k=0p1(1/(p1)k)p123p4Bp3 (modp5),\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 aZa\in\mathbb Z with pap\nmid a we have k=1p11k(1+ak)k1(modp)   \mboxand   k=1p11k2(1+ak)k1+12a(modp).\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}.

Keywords

Cite

@article{arxiv.1011.3487,
  title  = {Supercongruences motivated by e},
  author = {Zhi-Wei Sun},
  journal= {arXiv preprint arXiv:1011.3487},
  year   = {2015}
}

Comments

16 pages

R2 v1 2026-06-21T16:44:07.106Z