Combinatorial congruences modulo prime powers
| dc.creator | Sun, Zhi-Wei | |
| dc.creator | Davis, Donald M. | |
| dc.date | 2005-08-04 | |
| dc.date | 2007-07-25 | |
| dc.date.accessioned | 2026-07-07T08:20:01Z | |
| dc.date.available | 2026-07-07T08:20:01Z | |
| dc.description | Let p be any prime, and let a and n be nonnegative integers. Let $r\in Z$ and $f(x)\in Z[x]$. We establish the congruence $$p^{°f}\sum_{k=r(mod p^a)}\binom{n}{k}(-1)^k f((k-r)/p^a) =0 (mod p^{\sum_{i=a}^{\infty}[n/p^i]})$$ (motivated by a conjecture arising from algebraic topology), and obtain the following vast generalization of Lucas' theorem: If a is greater than one, and $l,s,t$ are nonnegative integers with $s,t<p$, then $$\frac{1}{[n/p^{a-1}]!} \sum_{k=r(mod p^a)} \binom{pn+s}{pk+t}(-1)^{pk}((k-r)/p^{a-1})^l =\frac {1}{[n/p^{a-1}]!} \sum_{k=r(mod p^a)}\binom{n}{k}\binom{s}{t}(-1)^k((k-r)/p^{a-1})^l (mod p).$$ We also present an application of the first congruence to Bernoulli polynomials, and apply the second congruence to show that a p-adic order bound given by the authors in a previous paper can be attained when p=2. | |
| dc.identifier | https://arxiv.org/abs/math/0508087 | |
| dc.identifier | http://arxiv.org/abs/math/0508087 | |
| dc.identifier | Trans. Amer. Math. Soc. 359(2007), no.11, 5525-5553 | |
| dc.identifier.uri | http://salesiana.dossiersoluciones.com/handle/123456789/134966 | |
| dc.subject | Number Theory | |
| dc.subject | Combinatorics | |
| dc.subject | 11B65, 05A10, 11A07, 11B68, 11S05 | |
| dc.title | Combinatorial congruences modulo prime powers | |
| dc.type | text |