Related papers: Some supercongruences modulo $p^2$

In this paper we mainly employ the Zeilberger algorithm to study congruences for sums of terms involving products of three binomial coefficients. Let $p>3$ be a prime. We prove that…

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…

In this paper we prove the supercongruence $$\sum_{n=0}^{(p-1)/2}\frac{6n+1}{256^n}\binom{2n}n^3\equiv p(-1)^{(p-1)/2}+(-1)^{(p-1)/2}\frac{7}{24}p^4B_{p-3}\pmod{p^5}$$ for any prime $p>3$, which was conjectured by Sun in 2019.

In this paper, we confirm several conjectured congruences of Sun concerning the divisibility of binomial sums. For example, with help of a quadratic hypergeometric transformation, we prove that $$…

The polynomials $d_n(x)$ are defined by \begin{align*} d_n(x) &= \sum_{k=0}^n{n\choose k}{x\choose k}2^k. \end{align*} We prove that, for any prime $p$, the following congruences hold modulo $p$: \begin{align*}…

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…

The numbers $R_n$ and $W_n$ are defined as \begin{align*} R_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{1}{2k-1},\ \text{and}\ W_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{3}{2k-3}. \end{align*} We prove that, for any positive…

Recently, Z.-W. Sun made the following conjecture: for any odd prime $p$ and odd integer $m$, $$ \frac{1}{m^2{m-1\choose (m-1)/2}}\Bigg(\sum_{k=0}^{(pm-1)/2}\frac{{2k\choose k}}{8^k}…

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 $p>3$, $$…

Let $M_n$ and $T_n$ denote the $n$th Motzkin number and the $n$th central trinomial coefficient respectively. We prove that for any prime $p\ge 5$, \begin{align*} &\sum_{k=0}^{p-1}M_k^2\equiv…

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…

Let $p>3$ be a prime, and let $a$ be a rational p-adic integer with $a\not\equiv 0\pmod p$. In this paper we establish congruences for $$\sum_{k=1}^{(p-1)/2}\frac{\binom ak\binom{-1-a}k}k, \quad\sum_{k=0}^{(p-1)/2}k\binom ak\binom{-1-a}k…

Let $m>2$ and $q>0$ be integers with $m$ even or $q$ odd. We show the supercongruence $$\sum_{k=0}^{p-1}(-1)^{km}\binom{p/m-q}{k}^m\equiv0\pmod{p^3}.$$ for any prime $p>mq$. This confirms a conjecture of Sun.

In this paper, we mainly prove the following conjectures of Z.-H. Sun \cite{SH2}: Let $p>3$ be a prime. If $p\equiv1\pmod3$ and $p=x^2+3y^2$, then we have $$…

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

In this paper, we mainly prove the following conjectures of Z.-W. Sun \cite{S13}: Let $p>2$ be a prime. If $p=x^2+3y^2$ with $x,y\in\mathbb{Z}$ and $x\equiv1\pmod 3$, then $$x\equiv\frac14\sum_{k=0}^{p-1}(3k+4)\frac{f_k}…

In this paper, we employ the Wilf-Zeilberger (WZ) method to prove a supercongruence conjecture posed by Z.-W. Sun: for any prime $p$, \begin{align*} \sum_{k=0}^{\frac{p-3}{2}}\frac{92k^2+61k+9}{(2k+1)64^k}{2k \choose k}{3k \choose k}{4k…

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/k$ be the $n$th harmonic number. Z. W. Sun conjectured that for any prime $p\ge 5$, $$ \sum_{k=1}^{p-1}\frac{H_k}{k\cdot 2^k} \equiv7/24pB_{p-3}\pmod{p^2}. $$ This conjecture is recently…

Let p be a prime and let a be a positive integer. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=1}^{p-1}\binom{2k}{k+d}/(km^{k-1})$ modulo $p$ for all d=0,...,p^a, where m is any integer not divisible by p.…

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…