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About von Staudt congruences for Bernoulli numbers. (English) Zbl 1024.11011

From the text: The author considers the von Staudt type congruences for Bernoulli numbers ${B}_{n}$ with arbitrary indices $n$ (the case $n\equiv 0$ $p\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}-1$ being no exception). The theorems proved generalize well-known results due to H. S. Vandiver, L. Carlitz and others.

Theorem 1. Let $p$ be an odd prime, $n=k\left(p-1\right){p}^{l-1}$, $k$ and $l\in ℕ$. Then $p{B}_{n}\equiv p-1\phantom{\rule{4.44443pt}{0ex}}\left(mod\phantom{\rule{0.277778em}{0ex}}{p}^{l}\right)$ or more exactly

$p{B}_{n}\equiv p-1+k{p}^{l}\sum _{a=1}^{p-1}\left({a}^{p-1}-1\right)/p\phantom{\rule{10.0pt}{0ex}}\left(mod\phantom{\rule{0.277778em}{0ex}}{p}^{l+1}\right),\phantom{\rule{4pt}{0ex}}p>3·$

Theorem 2. Let $k,l,t,u\in ℕ$ and $z\in ℕ\cup \left\{0\right\}$. Then

${\delta }^{t+k-1}{B}^{hz+u}\equiv 0\phantom{\rule{10.0pt}{0ex}}\left(mod\phantom{\rule{0.277778em}{0ex}}{p}^{lt}\right),$

where $h=\varphi \left({p}^{l}\right)=\left(p-1\right){p}^{l-1}$, ${p}^{k}\ge \left(2k+lt\right)\left(p-1\right)+1$ and $u\ge l\left(t+k-1\right)$. In particular, for $k=2$ we have

${\delta }^{t+1}{B}^{hz+u}\equiv 0\phantom{\rule{10.0pt}{0ex}}\left(mod\phantom{\rule{0.277778em}{0ex}}{p}^{lt}\right),$

or (in the usual symbolic form)

${B}^{hz+u}{\left({B}^{h}-1\right)}^{t+1}\equiv 0\phantom{\rule{10.0pt}{0ex}}\left(mod\phantom{\rule{0.277778em}{0ex}}{p}^{lt}\right),$

where $p\ge lt+3$ and $u\ge l\left(t+1\right)$. ${B}_{n}$ denotes the Bernoulli number.

##### MSC:
 11B68 Bernoulli and Euler numbers and polynomials 11A07 Congruences; primitive roots; residue systems