中文

Fleck quotients and Bernoulli numbers

数论 2007-05-23 v1 组合数学

摘要

Let p be a prime, and let n>0 and r be integers. In 1913 Fleck showed that Fp(n,r)=(p)[(n1)/(p1)]k=r(modp)(nk)(1)kZ.F_p(n,r)=(-p)^{-[(n-1)/(p-1)]}\sum_{k=r(mod p)}\binom{n}{k}(-1)^k\in\Z. Nowadays this result plays important roles in many aspects. Recently Sun and Wan investigated Fp(n,r)F_p(n,r) mod p in [SW2]. In this paper, using p-adic methods we determine (Fp(m,r)Fp(n,r))/(mn)(F_p(m,r)-F_p(n,r))/(m-n) modulo p in terms of Bernoulli numbers, where m>0 is an integer with mnm\not=n and m=n(modp(p1))m=n (mod p(p-1)). Consequently, Fp(n,r)F_p(n,r) mod pordp(n)+1p^{ord_p(n)+1} is determined; for example, if n=n(modp1)n=n_*(mod p-1) with 0<n<p20<n_*<p-2 then Fp(pn,0)pn=n!n+1Bp1n(modp).\frac{F_p(pn,0)}{pn}=\frac{n_*!}{n_*+1}B_{p-1-n_*} (mod p). This yields an application to Stirling numbers of the second kind. We also study extended Fleck quotients; in particular we prove that if a>0a>0 and l0l\ge 0 are integers with 2nlp2\le n-l\le p then 1pnll<kn(pandpakd)(1)pk(k1l)=(1)l1n!l!(nl)Bpn+l(modp)\frac{1}{p^{n-l}}\sum_{l<k\le n} \binom{p^a n-d}{p^a k-d}(-1)^{pk}\binom{k-1}{l} =\frac{(-1)^{l-1}n!}{l!(n-l)}B_{p-n+l} (mod p) for all d=1,...,max{p^{a-2},1}.

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引用

@article{arxiv.math/0608328,
  title  = {Fleck quotients and Bernoulli numbers},
  author = {Zhi-Wei Sun},
  journal= {arXiv preprint arXiv:math/0608328},
  year   = {2007}
}

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38 pages