Related papers: New congruences for central binomial coefficients
We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd}…
Binomial coefficients and central trinomial coefficients play important roles in combinatorics. Let $p>3$ be a prime. We show that $$T_{p-1}\equiv\left(\frac p3\right)3^{p-1}\ \pmod{p^2},$$ where the central trinomial coefficient $T_n$ is…
Recently, using modular forms F. Beukers posed a unified method that can deal with a large number of supercongruences involving binomial coefficients and Ap\'ery-like numbers. In this paper, we use Beukers' method to prove some conjectures…
Let $p$ be an odd prime, Jianqiang Zhao has established a curious congruence $$ \sum_{i+j+k=p \atop i,j,k > 0} \frac{1}{ijk} \equiv -2B_{p-3}\pmod p , $$ where $B_{n}$ denotes the $n-$th Bernoulli numbers. In this paper, we will generalize…
Let p be any prime, and let a and n be nonnegative integers. Let $r\in Z$ and $f(x)\in Z[x]$. We establish the congruence $$p^{\deg 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]})$$ (motivated by…
Let $p>3$ be a prime, and let $m$ be an integer with $p\nmid m$. In the paper we prove some supercongruences concerning $$\align &\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom{3k}k}{54^k},\…
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 $p>2$ $$ \sum_{\substack{i+j+k=p^r\\ i,j,k\in{\mathfrak P}_p}}…
Let $p$ be an odd prime, and let $a$ be a rational $p$-adic integer with $a\not\equiv 0\pmod p$. In this paper, using WZ method we establish the congruences for $\sum_{k=0}^{p-1} \binom ak^2(-1)^k(1-\frac 2ak)$ modulo $p^2$ and…
We present some congruences modulo $p^{6-d}$ for sums of the type $\sum_{k=0}^{(p-3)/2}x^k{2k\choose k}/(2k+1)^d$, for $d=1,2,3$ where $p>5$ is a prime.
Let $p$ be an odd prime and let $d\in\{2,3,7\}$. When $(\frac{-d}p)=1$ we can write $p=x^2+dy^2$ with $x,y\in\mathbb Z$; in this paper we aim at determining $x$ or $y$ modulo $p^2$. For example, when $p=x^2+3y^2$, we show that if $p\equiv…
In this paper, we confirm some congruences conjectured by V.J.W. Guo and M.J. Schlosser recently. For example, we show that for primes $p>3$, $$…
In this paper, we partly prove a supercongruence conjectured by Z.-W. Sun in 2013. Let $p$ be an odd prime and let $a\in\mathbb{Z}^{+}$. Then if $p\equiv1\pmod3$, we have \begin{align*}…
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:…
The Ap\'ery polynomials are given by $$A_n(x)=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2x^k\ \ (n=0,1,2,\ldots).$$ (Those $A_n=A_n(1)$ are Ap\'ery numbers.) Let $p$ be an odd prime. We show that…
By a very simple argument, we prove that if $l,m,n$ are nonnegative integers then $$\sum_{k=0}^l(-1)^{m-k}\binom{l}{k}\binom{m-k}{n}\binom{2k}{k-2l+m} =\sum_{k=0}^l\binom{l}{k}\binom{2k}{n}\binom{n-l}{m+n-3k-l}. On the basis of this…
We give elementary proofs for the Apagodu-Zeilberger-Stanton-Amdeberhan-Tauraso congruences $$\sum\limits_{n=0}^{p-1}\dbinom{2n}{n} \equiv\eta_{p}\mod p^{2},$$ $$\sum\limits_{n=0}^{rp-1}\dbinom{2n}{n}…
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…
Let $m$, $r$ and $n$ be positive integers. We denote by ${\bf k}\vdash n$ any tuple of odd positive integers ${\bf k}=(k_1,\dots,k_t)$ such that $k_1+\dots+k_t=n$ and $k_j\ge 3$ for all $j$. In this paper we prove that for every…
Let k>1 be an integer and let p be a prime. We show that if $p^a\le k<2p^a$ or $k=p^aq+1$ (with 2q<p) for some a=1,2,..., then the set {\binom{n}{k}: n=0,1,2,...} is dense in the ring Z_p of p-adic integers, i.e., it contains a complete…
Let $p$ be an odd prime. In 2008 E. Mortenson proved van Hamme's following conjecture: $$\sum_{k=0}^{(p-1)/2}(4k+1)\binom{-1/2}k^3\equiv (-1)^{(p-1)/2}p\pmod{p^3}.$$ In this paper we show further that…