English

Super congruences involving Bernoulli and Euler polynomials

Number Theory 2014-07-29 v6

Abstract

Let p>3p>3 be a prime, and let aa be a rational p-adic integer. Let {Bn(x)}\{B_n(x)\} and {En(x)}\{E_n(x)\} denote the Bernoulli polynomials and Euler polynomials, respectively. In this paper we show that k=0p1(ak)(1ak)(1)ap+p2t(t+1)Ep3(a)(modp3)\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≢12(modp)a\not\equiv -\frac 12\pmod p, k=0p1(ak)(1ak)12k+11+2t1+2a+p2t(t+1)1+2aBp2(a)(modp3),\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 ap{0,1,,p1}\langle a\rangle_p\in\{0,1,\ldots,p-1\} satisfying aap(modp)a\equiv \langle a\rangle_p\pmod p and t=(aap)/pt=(a-\langle a\rangle_p)/p. Taking a=13,14,16a=-\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 k=0p1k(ak)(1ak), k=0p1(ak)(1ak)12k1, k=1p11k(ak)(1ak)(modp3)\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 k=1p1(1)kk(ak), k=0p1(ak)(2)k(modp2).\sum_{k=1}^{p-1}\frac {(-1)^k}k\binom ak,\ \sum_{k=0}^{p-1}\binom ak(-2)^k\pmod{p^2}.

Keywords

Cite

@article{arxiv.1407.0636,
  title  = {Super congruences involving Bernoulli and Euler polynomials},
  author = {Zhi-Hong Sun},
  journal= {arXiv preprint arXiv:1407.0636},
  year   = {2014}
}

Comments

30 pages

R2 v1 2026-06-22T04:53:37.267Z