相关论文: Congruences involving $\binom{4k}{2k}$ and $\binom…

In this paper, we establish the following two congruences: \begin{gather*} \sum_{k=0}^{(p+1)/2}(3k-1)\frac{\left(-\frac{1}{2}\right)_k^2\left(\frac{1}{2}\right)_k4^k}{k!^3}\equiv…

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.

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}}…

For n=1,2,3,... let p_n be the n-th prime. We mainly show that p_n>n+sum_{k=1}^n p_k/k for all n>124, and sum_{k=1}^n kp_k<n^2p_n/3 for all n>30.

A prime number $p$ is said to be a Wolstenholme prime if it satisfies the congruence ${2p-1\choose p-1} \equiv 1 \,\,(\bmod{\,\,p^4})$. For such a prime $p$, we establish the expression for ${2p-1\choose p-1}\,\,(\bmod{\,\,p^8})$ given in…

Let $p$ be an odd prime and let $a,m$ be integers with $a>0$ and $m \not\equiv0\pmod p$. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ mod $p^2$ for $d=0,1$; for example,…

Define $g_n(x)=\sum_{k=0}^n\binom nk^2\binom{2k}kx^k$ for $n=0,1,2,...$. Those numbers $g_n=g_n(1)$ are closely related to Ap\'ery numbers and Franel numbers. In this paper we establish some fundamental congruences involving $g_n(x)$. For…

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…

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.

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…

Let $[x]$ be the greatest integer not exceeding $x$. In the paper we introduce the sequence $\{U_n\}$ given by $U_0=1$ and $U_n=-2\sum_{k=1}^{[n/2]}\binom n{2k}U_{n-2k}\quad(n\ge 1)$, and establish many recursive formulas and congruences…

For $n\in\mathbb{N}=\{0,1,2,\ldots\}$ and $b,c\in\mathbb{Z}$, the $n$th generalized central trinomial coefficient $T_n(b,c)$ is the coefficient of $x^n$ in the expansion of $(x^2+bx+c)^n$. In particular, $T_n=T_n(1,1)$ is the central…

In this paper, we mainly prove the following conjecture of Z.-H. Sun cite{SH20}: Let $p>3$ be a prime. Then $$\sum_{k=0}^{p-1}\binom{2k}k\frac{3k+1}{(-16)^k}f_k\equiv(-1)^{(p-1)/2}p+p^3E_{p-3}\pmod{p^4},$$ where…

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 $m$, $r$ 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…

In this paper we establish some new supercongruences motivated by the well-known fact $\lim_{n\to\infty}(1+1/n)^n=e$. Let $p>3$ be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{-1/(p+1)}k^{p+1}\equiv 0\ \pmod{p^5}\ \ \ \mbox{and}\ \ \…

In this paper, we pose lots of challenging conjectures on congruences for the sums involving binomial coefficients and Ap\'ery-like numbers modulo $p^3$, where $p$ is an odd prime.

In 2017, motivated by a supercongruence conjectured by Kimoto and Wakayama and confirmed by Long, Osburn and Swisher, Z.-W. Sun introduced the sequence of polynomials: $$…

Kemnitz Conjecture [9] states that if we take a sequence of elements in $Z_{p}^{2}$ of length $4p-3$, $p$ is a prime number, then it has a subsequence of length $p$, whose sum is $0$ modulo $p$. It is known that in $Z_{p}^{3}$ to get a…

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…