Related papers: Proof of a conjecture of Adamchuk

Let $p$ be a prime with $p>3$, and let $a,b$ be two rational $p-$integers. In this paper we present general congruences for $\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac p{k+b}\pmod {p^2}$. For $n=0,1,2,\ldots$ let $D_n$ and $b_n$ be Domb…

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…

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…

For various positive integers $k$, the sums of $k$th powers of the first $n$ positive integers, $S_k(n+1)=1^k+2^k+...+n^k$, have got to be some of the most popular sums in all of mathematics. In this note we prove that for each $k\ge 2 $$…

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 $p>5$ 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…

In this paper, we prove two conjectures of Z.-W. Sun: $$2n\binom{2n}n\big|\sum_{k=0}^{n-1}(3k+1)\binom{2k}k^3{16}^{n-1-k}\ \mbox{for}\ \mbox{all}\ n=2,3,\cdots,$$ and $$\sum_{k=0}^{(p-1)/2}\frac{3k+1}{16^k}\binom{2k}{k}^3\equiv…

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. In the paper, by using the properties of Legendre polynomials we prove some congruences for $\sum_{k=0}^{\frac{p-1}2}\binom{2k}k^2m^{-k}\mod {p^2}$. In particular, we confirm several conjectures of Z.W. Sun. We also…

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…

Let p be an odd prime and let a be a positive integer. In this paper we investigate the sum $\sum_{k=0}^{p^a-1}\binom{hp^a-1}{k}\binom{2k}{k}/m^k$ mod p^2, where h,m are p-adic integers with m\not=0 (mod p). For example, we show that if…

Employing the $q$-Lucas theorem and some known $q$-supercongruences, we give some Dwork-type $q$-congruences, confirming three conjectures in [J. Combin. Theory, Ser. A 178 (2021), Art.~105362]. As conclusions, we obtain the following…

It is significant to study congruences involving multiple harmonic sums. Let $p$ be an odd prime, in recent years, the following curious congruence $$\sum_{\substack{i+j+k=p \\ i, j, k>0}} \frac{1}{i j k} \equiv-2 B_{p-3}\pmod p$$ has been…

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…

We prove congruences, modulo a power of a prime p, for certain finite sums involving central binomial coefficients $\binom{2k}{k}$.

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…

For a non-negative integer $m$, let $S(m)$ denote the sum given by $$S(m):=\sum_{n=0}^{m}\frac{(-1)^n(8n+1)}{n!^3}\left(\frac{1}{4}\right)_n^3.$$ Using the powerful WZ-method, for a prime $p\equiv 3$ $($mod $4)$ and an odd integer $r>1$, we…

Let $p\equiv3\pmod{4}$ be a prime and $r$ a positive integer. We show that $$ \prod_{k=1}^{(p^{2r}-1)/2}\frac{4k-1}{4k+1}\equiv1\pmod{p^2}. $$ This confirms a recent conjecture of Guo.

For any odd prime p we obtain q-analogues of Van Hamme's supercongruence: $$ \sum_{k=0}^{\frac{p-1}{2}}{2k\choose k}^3\frac{1}{64^k} \equiv 0 \pmod{p^2} \quad\text{for}\quad p\equiv 3\pmod 4, $$ and Rodriguez-Villegas' Beukers-like…