Related papers: Supercongruences involving binomial coefficients a…

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 $p\ge 5$ and $r\ge 2$, we prove that \begin{equation} \sum\limits_{\begin{smallmatrix}…

We establish a parametric supercongruence related to Whipple's ${}_5F_4$ formula and Dwork's dash operation. As a typical consequence, we obtain the following result: for any prime $p\equiv3\pmod4$ and odd integer $r\geq1$, $$…

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 prove that if $q$ is a power of a prime $p$ and $p^k$ divides $a$, with $k\ge 0$, then \[ 1+(q-1)\sum_{0\le b(q-1)<a} \binom{a}{b(q-1)}\equiv 0\pmod{p^{k+1}}. \] The special case of this congruence where $q=p$ was proved by Carlitz in…

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 $$…

A classical result of A. Fleck states that if p is a prime, and n>0 and r are integers, then $$\sum_{k=r(mod p)}\binom {n}{k}(-1)^k=0 (mod p^{[(n-1)/(p-1)]}).$$ Recently R. M. Wilson used Fleck's congruence and Weisman's extension to…

We give a simple matrix-based proof of congruence equations modulo a prime $p$ involving sums of binomial coefficients appearing in Pascal's triangle. These equations can be used to construct some groups of exponent $p^n$. These groups, as…

In $2012$, Guillera and Zudilin established the following two supercongruences involving truncated Ramanujan-type series: for any odd prime $p>2$, \begin{align*}…

In 2014, Wang and Cai established the following harmonic congruence for any odd prime $p$ and positive integer $r$, \begin{equation*} Z(p^{r})\equiv-2p^{r-1}B_{p-3} ~(\bmod ~ p^{r}), \end{equation*} where $…

The following congruence for power sums, $S_n(p)$, is well known and has many applications: $1^n+2^n +\dots +p^n \equiv\begin{cases} -1 \text{ mod } p, & \text{ if } \ p-1 \ | \ n; 0 \text{ mod } p, & \text{ if } \ p-1 \ \not| \ n,…

We prove that $\sum_{k=0}^{q-1}\binom{2k}{k}\equiv q^2\pmod{3q^2}$ if q>1 is a power of 3, as recently conjectured by Z.W. Sun and R. Tauraso. Our more precise result actually implies that the value of $(1/q^2)\sum_{k=0}^{q-1}\binom{2k}{k}$…

By examining asymptotic behavior of certain infinite basic ($q$-) hypergeometric sums at roots of unity (that is, at a "$q$-microscopic" level) we prove polynomial congruences for their truncations. The latter reduce to non-trivial…

Let $\overline{p}_{k}(n)$ denote the number of overpartition $k$-tuples of $n$. In 2023, Saikia \cite{saikia} conjectured the following congruences: \begin{align*} \overline{p}_{q}(8n+2)& \equiv 0 \pmod{4},\quad \overline{p}_{q}(8n+3)\equiv…

Let $p$ be any prime and let $a$ and $n$ be positive integers with $p\nmid n$. We show that $$\sum_{k=1}^{p^a-1}\frac{B_k}{(-n)^k}\equiv a(-1)^{n-1}D_{n-1}\pmod {p},$$ where $B_0,B_1,\ldots$ are the Bell numbers and $D_0,D_1,\ldots$ are the…

Let $p$ be an odd prime and let $d$ be an integer not divisible by $p$. We prove that $$ \prod_{1\le m,n\le p-1\atop p\nmid m^2-dn^2}\ (x-(m+n\sqrt{d})) \equiv \begin{cases}\sum_{k=1}^{p-2}\frac{k(k+1)}2x^{(k-1)(p-1)}\pmod p &\text{if}\…

We evaluate some series with summands involving a single binomial coefficient $\binom{6k}{3k}$. For example, we prove that $$\sum_{k=0}^\infty\frac{(63k^2+78k+22)8^k}{(2k+1)(6k+1)(6k+5)\binom{6k}{3k}}=\frac{3\pi}2.$$ Motivated by Galois…

With help of $q$-congruence, we prove the divisibility of some binomial sums. For example, for any integers $\rho,n\geq 2$, $$\sum_{k=0}^{n-1}(4k+1) \binom{2k}{k}^\rho \cdot (-4)^{\rho(n-1-k)} \equiv 0\pmod{2^{\rho-2}n\binom{2n}{n}}.$$

In this paper, we prove a congruence which confirms a conjecture of Adamchuk. For any prime $p\equiv1\pmod3$ and $a\in\mathbb{Z}^{+}$, we have \begin{align*} \sum_{k=1}^{\frac{2}3(p^a-1)}\binom{2k}k\equiv0\pmod{p^2}. \end{align*}

The purpose of this note is to obtain some congruences modulo a power of a prime $p$ involving the truncated hypergeometric series $$\sum_{k=1}^{p-1} {(x)_k(1-x)_k\over (1)_k^2}\cdot{1\over k^a}$$ for $a=1$ and $a=2$. In the last section,…

We show that for any prime prime $p\not=2$ $$\sum_{k=1}^{p-1} {(-1)^k\over k}{-{1\over 2} \choose k} \equiv -\sum_{k=1}^{(p-1)/2}{1\over k} \pmod{p^3}$$ by expressing the l.h.s. as a combination of alternating multiple harmonic sums.