Related papers: A Curious Congruence Involving Alternating Harmoni…

We determine the explicit formulas for the sum of products of homogeneous multiple harmonic sums $\sum_{k=1}^n \prod_{j=1}^r H_k(\{1\}^{\lambda_j})$ when $\sum_{j=1}^r \lambda_j\leq 5$. We apply these identities to the study of two…

In this paper we study a family of polynomials $$S_n^{(m)}(x):=\sum_{i,j=0}^n\binom ni^m\binom nj^m\binom{i+j}ix^{i+j}\ \ (m,n=0,1,2,\ldots).$$ For example, we show that $$\sum_{k=0}^{p-1}S_k^{(0)}(x)\equiv\frac…

In this work we consider the congruence $\sum_{j=1}^{n-1} j^{k(n-1)} \equiv -1 \pmod n$ for each $k \in \mathbb{N}$, thus extending Giuga's ideas for $k=1$. In particular, it is proved that a pair $(n,k)\in \mathbb{N}^2$ satisfies this…

In this paper, we investigate some congruences involving sums of $\frac{d^{-k}{x\choose k}{x+k\choose k}}{{2k \choose k}}$, where $x$ be a $p$-adic integer, $k$ be a non-negative integer, and $d$ $(d\neq 0)$ be a rational number.

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that $\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ and $\nu_p(\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{\binom{2k}k}{m^k})$ are at…

We prove some supercongruence and divisibility results on sums involving Domb numbers, which confirm four conjectures of Z.-W. Sun and Z.-H. Sun. For instance, by using a transformation formula due to Chan and Zudilin, we show that for any…

Let $k\geq 2$ be a fixed natural number. We establish the existence of infinitely many pairs of consecutive primes $p_n$, $p_{n+1}$ satisfying $$ p_{n+1}-p_n\geq c\:\frac{\log p_n\: \log_2 p_n\: \log_4 p_n}{\log_3 p_n}\:,$$ with $c$ being a…

Let $n, k$ and $a$ be positive integers. The Stirling numbers of the first kind, denoted by $s(n,k)$, count the number of permutations of $n$ elements with $k$ disjoint cycles. Let $p$ be a prime. In recent years, Lengyel, Komatsu and…

Let p be any odd prime. We mainly show that $$\sum_{k=1}^{p-1}binomial(3k,k)*2^k/k=0 (mod p)$$ and $$\sum_{k=1}^{p-1}2^{k-1}C_k^{(2)}=(-1)^{(p-1)/2}-1 (mod p),$$ where $C_k^{(2)}=binomial(3k,k)/(2k+1)$ is the $k$th Catalan number of order…

In this paper, we prove some supercongruences concerning truncated hypergeometric series. For example, we show that for any prime $p>3$ and positive integer $r$, $$ \sum_{k=0}^{p^r-1}(3k+1)\frac{(\frac12)_k^3}{(1)_k^3}4^k\equiv…

For any measure preserving system $(X,\mathcal{X},\mu,T)$ and $A\in\mathcal{X}$ with $\mu(A)>0$, we show that there exist infinitely many primes $p$ such that $\mu\bigl(A\cap T^{-(p-1)}A\cap T^{-2(p-1)}A\bigr) > 0$ (the same holds with…

If p is a prime and n a positive integer, let v(n) denote the exponent of p in n, and u(n)=n/p^{v(n)} the unit part of n. If k is a positive integer not divisible by p, we show that the p-adic limit of (-1)^{pke} u((kp^e)!) as e goes to…

We show that if p is an odd prime then $$\sum_{k=0}^{p-1}E_kE_{p-1-k}=1 (mod p)$$ and $$\sum_{k=0}^{p-3}E_kE_{p-3-k}=(-1)^{(p-1)/2}2E_{p-3} (mod p),$$ where E_0,E_1,E_2,... are Euler numbers. Moreover, we prove that for any positive integer…

In this paper, we apply the power-partible reduction to study arithmetic properties of sums involving Delannoy numbers $D_k$ and polynomials $D_k(z)$. Let $v\in\bN$ and $p$ be an odd prime. It is proved that, for any…

Many generating series of combinatorially interesting numbers have the property that the sum of the terms of order $<p$ at some suitable point is congruent to a zero of a zeta-function modulo infinitely many primes $p$. Surprisingly, very…

It is well-known that the congruence $\sum_{i=1}^{ n} i^{ n} \equiv 1 \pmod{n}$ has exactly five solutions: $\{1,2,6,42,1806\}$. In this work, we characterize the solutions to the congruence $1^n + 2^n + \dotsb + n^n\equiv p \pmod{n}$ for…

For n=1,2,3,... let p_n be the n-th prime. We mainly show that p_n>n+sum_{k=1}^n p_k/k for all n>124, and sum_{k=1}^n kp_k<n^2p_n/3 for all n>30.

Let $p$ be a prime. In 1878 \'{E}. Lucas proved that the congruence $$ {p-1\choose k}\equiv (-1)^k\pmod{p}$$ holds for any nonnegative integer $k\in\{0,1,\ldots,p-1\}$. The converse statement was given in Problem 1494 of {\it Mathematics…

By a very simple argument, we prove that if $l,m,n$ are nonnegative integers then $$\sum_{k=0}^l(-1)^{m-k}\binom{l}{k}\binom{m-k}{n}\binom{2k}{k-2l+m} =\sum_{k=0}^l\binom{l}{k}\binom{2k}{n}\binom{n-l}{m+n-3k-l}. On the basis of this…

Let $p>5$ be a prime. Motivated by the known formulae $\sum_{k=1}^\infty(-1)^k/(k^3\binom{2k}{k})=-2\zeta(3)/5$ and $\sum_{k=0}^\infty \binom{2k}{k}^2/((2k+1)16^k)=4G/\pi$$ (where $G=\sum_{k=0}^\infty(-1)^k/(2k+1)^2$ is the Catalan…