Supercongruences involving Euler polynomials
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Abstract:
Let $p>3$ be a prime, and let $a$ be a rational $p$-adic integer. Let $\{E_n(x)\}$ denote the Euler polynomials given by $\frac {2\text {e}^{xt}}{\text {e}^t+1}=\sum _{n=0}^{\infty }E_n(x)\frac {t^n}{n!}$. In this paper we show that \begin{align*} &\sum _{k=0}^{p-1}\binom ak\binom {-1-a}k\equiv (-1)^{\langle a\rangle _p}+ (a-\langle a\rangle _p)(p+a-\langle a\rangle _p)E_{p-3}(-a)\pmod {p^3}, \\&\sum _{k=0}^{p-1}\binom ak(-2)^k\equiv (-1)^{\langle a\rangle _p}-(a-\langle a\rangle _p)E_{p-2}(-a) \pmod {p^2}\quad \text {for}\quad a\not \equiv 0\pmod p, \end{align*} where $\langle a\rangle _p\in \{0,1,\ldots ,p-1\}$ satisfying $a\equiv \langle a\rangle _p\pmod p$. Taking $a=-\frac 13,-\frac 14,-\frac 16$ in the first congruence, we solve some conjectures of Z. W. Sun. We also establish a congruence for $\sum _{k=0}^{p-1}k\binom ak\binom {-1-a}k$ modulo $p^3$.References
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Additional Information
- Zhi-Hong Sun
- Affiliation: School of Mathematical Sciences, Huaiyin Normal University, Huaian, Jiangsu 223001, People’s Republic of China
- MR Author ID: 318137
- Email: zhihongsun@yahoo.com
- Received by editor(s): November 27, 2014
- Received by editor(s) in revised form: October 13, 2015
- Published electronically: February 2, 2016
- Additional Notes: The author was supported by the National Natural Science Foundation of China (grant No. 11371163).
- Communicated by: Matthew A. Papanikolas
- © Copyright 2016 American Mathematical Society
- Journal: Proc. Amer. Math. Soc. 144 (2016), 3295-3308
- MSC (2010): Primary 11A07; Secondary 11B68, 05A19
- DOI: https://doi.org/10.1090/proc/13005
- MathSciNet review: 3503698