Related papers: Supercongruences on some binomial sums involving L…

In this paper we study a family of polynomials $$S_n^{(m)}(x):=\sum_{i,j=0}^n\binom ni^m\binom nj^m\binom{i+j}ix^{i+j}\ \ (m,n=0,1,2,\ldots).$$ For example, we show that $$\sum_{k=0}^{p-1}S_k^{(0)}(x)\equiv\frac…

Consider an ordinary generating function $\sum_{k=0}^{\infty}c_kx^k$, of an integer sequence of some combinatorial relevance, and assume that it admits a closed form $C(x)$. Various instances are known where the corresponding truncated sum…

We utilize the Wilf-Zeilberger (WZ) method to establish congruences related to truncated Ramanujan-type series. By constructing hypergeometric terms $f(k, a, b, \ldots)$ with Gosper-summable differences and selecting appropriate parameters,…

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…

Let $p>3$ be a prime, and let $q_p(2)=(2^{p-1}-1)/p$ be the Fermat quotient of $p$ to base 2. Recently, Z. H. Sun proved that \sum_{k=1}^{p-1}\frac{1}{k\cdot 2^k}\equiv q_p(2)-\frac{p}{2}q_p(2)^2 \pmod{p^2} which is a generalization of a…

Let p be any odd prime. We mainly show that $$\sum_{k=1}^{p-1}binomial(3k,k)*2^k/k=0 (mod p)$$ and $$\sum_{k=1}^{p-1}2^{k-1}C_k^{(2)}=(-1)^{(p-1)/2}-1 (mod p),$$ where $C_k^{(2)}=binomial(3k,k)/(2k+1)$ is the $k$th Catalan number of order…

Let $p^k m^2$ be an odd perfect number with special prime $p$. In this article, we provide an alternative proof for the biconditional that $\sigma(m^2) \equiv 1 \pmod 4$ holds if and only if $p \equiv k \pmod 8$. We then give an application…

We give q-analogues of the following congruences by Z.-W. Sun: \sum_{k=1}^{p-1}\frac{D_k}{k} \equiv -\frac{2^{p-1}-1}{p} \pmod p,\\ \sum_{k=1}^{p-1}\frac{H_k}{k 2^k}\equiv 0 \pmod{p},\quad p\geqslant 5, where p is a prime,…

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that $\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ and $\nu_p(\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{\binom{2k}k}{m^k})$ are at…

In this paper, we prove two conjectural supercongruences on the $(p-1)$th Ap\'ery number, which were recently proposed by Z.-H. Sun.

In the recent article arXiv:1606.03351, Apagodu and Zeilberger discuss some applications of an algorithm for finding and proving congruence identities (modulo primes) of indefinite sums of many combinatorial sequence. At the end they…

For all nonnegative integers n, the Franel numbers are defined as $$ f_n=\sum_{k=0}^n {n\choose k}^3.$$ We confirm two conjectures of Z.-W. Sun on congruences for Franel numbers: \sum_{k=0}^{n-1}(3k+2)(-1)^k f_k &\equiv 0 \pmod{2n^2},…

Recently, Z.-W. Sun introduced two kinds of polynomials related to the Delannoy numbers, and proved some supercongruences on sums involving those polynomials. We deduce new summation formulas for squares of those polynomials and use them to…

For any $n\in\mathbb{N}=\{0,1,2,\ldots\}$ and $b,c\in\mathbb{Z}$, the generalized central trinomial coefficient $T_n(b,c)$ denotes the coefficient of $x^n$ in the expansion of $(x^2+bx+c)^n$. Let $p$ be an odd prime. In this paper, we…

Using the following $_4F_3$ transformation formula $$ \sum_{k=0}^{n}{-x-1\choose k}^2{x\choose n-k}^2=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}^2{x+k\choose 2k}, $$ which can be proved by Zeilberger's algorithm, we confirm some special…

Let $p$ be a prime, and let $f(x)$ be an integer-valued polynomial. By a combinatorial approach, we obtain a nontrivial lower bound of the $p$-adic order of the sum $$\sum_{k=r(mod p^{\beta})}\binom{n}{k}(-1)^k f([(k-r)/p^{\alpha}]),$$…

Let $p$ be a prime greater than 3. In the paper we mainly determine $\sum_{k=0}^{[p/4]}\binom{4k}{2k}(-1)^k$, $\sum_{k=0}^{[p/3]}\binom{3k}k, \sum_{k=0}^{[p/3]}\binom{3k}k(-1)^k$ and $\sum_{k=0}^{[p/3]}\binom{3k}k(-3)^k$ modulo $p$, where…

Via symbolic summation method, we establish the following series for $\pi^2$: \begin{align*} \sum_{k=1}^\infty \frac{H_k-2H_{2k}}{(-3)^k k} = \frac{\pi^2}{18}, \end{align*} where $H_k=\sum_{j=1}^k 1/j$. We also derive a $p$-adic congruence…

We give a simplified presentation of some results about recurrences of certain sequences of binomial sums in terms of (generalized) Fibonacci and Lucas polynomials.

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…