Related papers: Supercongruences motivated by e

We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd}…

The Delannoy numbers and Schr\"oder numbers are given by \begin{align*} D_n=\sum_{k=0}^n{n\choose k}{n+k\choose k}\quad \text{and}\quad S_n=\sum_{k=0}^n{n\choose k}{n+k\choose k}\frac{1}{k+1}, \end{align*} respectively. Let $p>3$ be a…

Let $p$ be an odd prime, and let $m$ be an integer with $p\nmid m$. In this paper show that $$\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak\binom{-1-a}k}{m^k} \equiv 0\pmod p \quad\hbox{implies}\quad\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak…

In this paper, we prove several supercongruences conjectured by Z.-W. Sun ten years ago via certain strange hypergeometric identities. For example, for any prime $p>3$, we show that…

We give a family of congruences for the binomial coefficients ${kp-1\choose p-1}$ in terms of multiple harmonic sums, a generalization of the harmonic numbers. Each congruence in this family (which depends on an additional parameter $n$)…

For $k=1,2,\ldots$ let $H_k$ denote the harmonic number $\sum_{j=1}^k 1/j$. In this paper we establish some new congruences involving harmonic numbers. For example, we show that for any prime $p>3$ we have…

Let ${\mathcal{P}_{n}}$ denote the set of positive integers which are prime to $n$. Let $B_{n}$ be the $n$-th Bernoulli number. For any prime $p \ge 11$ and integer $r\ge 2$, we prove that $$ \sum\limits_{\begin{smallmatrix}…

In this paper, we confirm some congruences conjectured by V.J.W. Guo and M.J. Schlosser recently. For example, we show that for primes $p>3$, $$…

The harmonic numbers $H_n=\sum_{0<k\le n}1/k\ (n=0,1,2,\ldots)$ play important roles in mathematics. Let $p>3$ be a prime. With helps of some combinatorial identities, we establish the following two new congruences:…

Let $p$ be a prime and let $a$ be a positive integer. In this paper we investigate $\sum_{k=0}^{p^a-1}\binom[(h+1)k,k+d]/m^k$ modulo a prime $p$, where $d$ and $m$ are integers with $-h<d<=p^a$ and $m\not=0 (mod p)$. We also study…

Let $p$ be an odd prime, Jianqiang Zhao has established a curious congruence, which is $$ \sum_{i+j+k=p \atop i,j,k > 0} \frac{1}{ijk} \equiv -2B_{p-3}\pmod p , $$ where $B_{n}$ denotes the $n$-th Bernoulli number. In this paper, we will…

Let $p$ be a prime and ${\mathfrak P}_p$ the set of positive integers which are prime to $p$. Recently, Wang and Cai proved that for every positive integer $r$ and prime $p>2$ $$ \sum_{\substack{i+j+k=p^r\\ i,j,k\in{\mathfrak P}_p}}…

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…

Let $E_n$ be the $n$-th Euler number and $(a)_n=a(a+1)\cdots (a+n-1)$ the rising factorial. Let $p>3$ be a prime. In 2012, Sun proved the that $$ \sum^{(p-1)/2}_{k=0}(-1)^k(4k+1)\frac{(\frac{1}{2})_k^3}{k!^3} \equiv…

In this paper we prove that for any prime $p\ge 11$ holds $$ {2p-1\choose p-1}\equiv 1 -2p \sum_{k=1}^{p-1}\frac{1}{k} +4p^2\sum_{1\le i<j\le p-1}\frac{1}{ij}\pmod{p^7}. $$ This is a generalization of the famous Wolstenholme's theorem which…

In this paper, we establish the following two congruences: \begin{gather*} \sum_{k=0}^{(p+1)/2}(3k-1)\frac{\left(-\frac{1}{2}\right)_k^2\left(\frac{1}{2}\right)_k4^k}{k!^3}\equiv…

The Franel numbers are defined by $ f_n=\sum_{k=0}^n {n\choose k}^3. $ Motivated by the recent work of Z.-W. Sun on Franel numbers, we prove that \begin{align*} \sum_{k=0}^{n-1}(3k+1)(-16)^{n-k-1} {2k\choose k} f_k &\equiv…

Let $p$ be an odd prime, Jianqiang Zhao has established a curious congruence $$ \sum_{i+j+k=p \atop i,j,k > 0} \frac{1}{ijk} \equiv -2B_{p-3}\pmod p , $$ where $B_{n}$ denotes the $n-$th Bernoulli numbers. In this paper, we will generalize…

In this paper, we mainly prove a congruence conjecture of Z.-W. Sun \cite{Sjnt}: Let $p>5$ be a prime. Then $$ \sum_{k=(p+1)/2}^{p-1}\frac{\binom{2k}k^2}{k16^k}\equiv-\frac{21}2H_{p-1}\pmod{p^4}, $$ where $H_n$ denotes the $n$-th harmonic…

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…