Leudesdorf's theorem and Bernoulli numbers

I. Sh. Slavutsky

Displaying similar documents to “Leudesdorf's theorem and Bernoulli numbers”

A note on the congruence $(\binom{n p^{k}}{m p^{k}}) \equiv (\binom{n}{m}) (mod p^{r})$

Romeo Meštrović (2012)

Czechoslovak Mathematical Journal

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In the paper we discuss the following type congruences: $(\binom{n p^{k}}{m p^{k}}) \equiv (\binom{m}{n}) (mod p^{r}),$ where $p$ is a prime, $n$ , $m$ , $k$ and $r$ are various positive integers with $n \geq m \geq 1$ , $k \geq 1$ and $r \geq 1$ . Given positive integers $k$ and $r$ , denote by $W (k, r)$ the set of all primes $p$ such that the above congruence holds for every pair of integers $n \geq m \geq 1$ . Using Ljunggren’s and Jacobsthal’s type congruences, we establish several characterizations of sets $W (k, r)$ and inclusion relations between them for various values $k$ and $r$ . In particular, we prove that $W (k + i, r) = W (k - 1, r)$ for all $k \geq 2$ , $i \geq 0$ and...

A generalization of a necessary and sufficient condition for primality due to Vantieghem.

Kilford, L.J.P. (2004)

International Journal of Mathematics and Mathematical Sciences

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Congruences involving the Fermat quotient

Romeo Meštrović (2013)

Czechoslovak Mathematical Journal

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Let $p > 3$ be a prime, and let $q_{p} (2) = (2^{p - 1} - 1) / p$ be the Fermat quotient of $p$ to base $2$ . In this note we prove that $\sum_{k = 1}^{p - 1} \frac{1}{k \cdot 2^{k}} \equiv q_{p} (2) - \frac{p q_{p} {(2)}^{2}}{2} + \frac{p^{2} q_{p} {(2)}^{3}}{3} - \frac{7}{48} p^{2} B_{p - 3} (mod p^{3}),$ which is a generalization of a congruence due to Z. H. Sun. Our proof is based on certain combinatorial identities and congruences for some alternating harmonic sums. Combining the above congruence with two congruences by Z. H. Sun, we show that $q_{p} {(2)}^{3} \equiv - 3 \sum_{k = 1}^{p - 1} \frac{2^{k}}{k^{3}} + \frac{7}{16} \sum_{k = 1}^{(p - 1) / 2} \frac{1}{k^{3}} (mod p),$ which is just a result established by K. Dilcher and L. Skula. As another application, we obtain a congruence for the sum $\sum_{k = 1}^{p - 1} 1 / (k^{2} \cdot 2^{k})$ modulo $p^{2}$ that also generalizes...