arXiv 1407.06362014-07-28math.NT

涉及Bernoulli多项式和欧拉多项式的超同余

Super congruences involving Bernoulli and Euler polynomials

Zhi-Hong Sun

备注：: 30 pages

摘要：

令$p>3$是素数，并且令$a$是有理p-adic整数。令$\{B_n(x)\}$和$\{E_n(x)\}$分别表示伯努利多项式和欧拉多项式。在本文中，我们证明了$$\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\equiv (-1)^{\langle a\rangle_p}+ p^2t(t+1)E_{p-3}(-a)\pmod{p^3}$$和$a\not\equiv -\frac 12\pmod p$，$$\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac 1{2k+1}\equiv \frac{1+2t}{1+2a} +p^2\frac{t(t+1)}{1+2a}B_{p-2}(-a)\pmod{p^3},$$，其中$\langle a\rangle_p\in\{0,1,\ldots,p-1\}$满足$a\equiv \langle a\rangle_p\pmod p$和$t=(a-\langle a\rangle_p)/p$。取$a=-\frac 13,-\frac 14,-\frac 16$在上述同余式中，我们解出了Z.W.太阳本文还建立了$\sum_{k=0}^{p-1}k\binom ak\binom{-1-a}k,\ \sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac 1{2k-1},\ \sum_{k=1}^{p-1}\frac 1k\binom ak\binom{-1-a}k\pmod{p^3}$和$\sum_{k=1}^{p-1}\frac {(-1)^k}k\binom ak,\ \sum_{k=0}^{p-1}\binom ak(-2)^k\pmod{p^2}.$的同余式

英文摘要：

Let $p>3$ be a prime, and let $a$ be a rational p-adic integer. Let $\{B_n(x)\}$ and $\{E_n(x)\}$ denote the Bernoulli polynomials and Euler polynomials, respectively. In this paper we show that $$\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\equiv (-1)^{\langle a\rangle_p}+ p^2t(t+1)E_{p-3}(-a)\pmod{p^3}$$ and for $a\not\equiv -\frac 12\pmod p$, $$\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac 1{2k+1}\equiv \frac{1+2t}{1+2a} +p^2\frac{t(t+1)}{1+2a}B_{p-2}(-a)\pmod{p^3},$$ where $\langle a\rangle_p\in\{0,1,\ldots,p-1\}$ satisfying $a\equiv \langle a\rangle_p\pmod p$ and $t=(a-\langle a\rangle_p)/p$. Taking $a=-\frac 13,-\frac 14,-\frac 16$ in the above congruences we solve some conjectures of Z.W. Sun. In this paper we also establish congruences for $\sum_{k=0}^{p-1}k\binom ak\binom{-1-a}k,\ \sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac 1{2k-1},\ \sum_{k=1}^{p-1}\frac 1k\binom ak\binom{-1-a}k\pmod{p^3}$ and $\sum_{k=1}^{p-1}\frac {(-1)^k}k\binom ak,\ \sum_{k=0}^{p-1}\binom ak(-2)^k\pmod{p^2}.$

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

摘要：

英文摘要：

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