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Supercongruences involving Euler polynomials


Author: Zhi-Hong Sun
Journal: Proc. Amer. Math. Soc. 144 (2016), 3295-3308
MSC (2010): Primary 11A07; Secondary 11B68, 05A19
DOI: https://doi.org/10.1090/proc/13005
Published electronically: February 2, 2016
MathSciNet review: 3503698
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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

  $\displaystyle \sum _{k=0}^{p-1}\binom ak\binom {-1-a}k\equiv (-1)^{\langle a\rangle _p}+ (a-\langle a\rangle _p)(p+a-\langle a\rangle _p)E_{p-3}(-a)\pmod {p^3},$    
  $\displaystyle \sum _{k=0}^{p-1}\binom ak(-2)^k\equiv (-1)^{\langle a\rangle _p}-(a-\langle a\rangle _p)E_{p-2}(-a) \mod {p^2}$$\displaystyle \quad \text {for}\quad a\not \equiv 0\pmod p,$    

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$.

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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
Email: zhihongsun@yahoo.com

DOI: https://doi.org/10.1090/proc/13005
Keywords: Congruence, Euler number, Euler polynomial
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
Article copyright: © Copyright 2016 American Mathematical Society