English

Supercongruences involving dual sequences

Number Theory 2017-04-21 v5 Combinatorics

Abstract

In this paper we study some sophisticated supercongruences involving dual sequences. For n=0,1,2,n=0,1,2,\ldots define dn(x)=k=0n(nk)(xk)2kd_n(x)=\sum_{k=0}^n\binom nk\binom xk2^k and sn(x)=k=0n(nk)(xk)(x+kk)=k=0n(nk)(1)k(xk)(1xk).s_n(x)=\sum_{k=0}^n\binom nk\binom xk\binom{x+k}k=\sum_{k=0}^n\binom nk(-1)^k\binom xk\binom{-1-x}k. For any odd prime pp and pp-adic integer xx, we determine k=0p1(±1)kdk(x)2\sum_{k=0}^{p-1}(\pm1)^kd_k(x)^2 and k=0p1(2k+1)dk(x)2\sum_{k=0}^{p-1}(2k+1)d_k(x)^2 modulo p2p^2; for example, we establish the new pp-adic congruence k=0p1(1)kdk(x)2(1)xp(modp2),\sum_{k=0}^{p-1}(-1)^kd_k(x)^2\equiv(-1)^{\langle x\rangle_p}\pmod{p^2}, where xp\langle x\rangle_p denotes the least nonnegative integer rr with xr(modp)x\equiv r\pmod p. For any prime p>3p>3 and pp-adic integer xx, we determine k=0p1sk(x)2\sum_{k=0}^{p-1}s_k(x)^2 modulo p2p^2 (or p3p^3 if x{0,,p1}x\in\{0,\ldots,p-1\}), and show that k=0p1(2k+1)sk(x)20(modp2).\sum_{k=0}^{p-1}(2k+1)s_k(x)^2\equiv0\pmod{p^2}. We also pose several related conjectures.

Keywords

Cite

@article{arxiv.1512.00712,
  title  = {Supercongruences involving dual sequences},
  author = {Zhi-Wei Sun},
  journal= {arXiv preprint arXiv:1512.00712},
  year   = {2017}
}

Comments

35 pages, final published version

R2 v1 2026-06-22T11:59:38.927Z