Related papers: Three pairs of congruences concerning sums of cent…

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…

The harmonic numbers $H_n=\sum_{0<k\le n}1/k\ (n=0,1,2,\ldots)$ play important roles in mathematics. Let $p>3$ be a prime. With helps of some combinatorial identities, we establish the following two new congruences:…

Let $p>3$ be a prime. For any $p$-adic integer $a$, we determine $$\sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k,\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k^{(2)},\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}k\frac{H_k^{(2)}}{2k+1}$$ modulo…

We give a $q$-congruence whose specializations $q=-1$ and $q=1$ correspond to supercongruences (B.2) and (H.2) on Van Hamme's 1997 list: $$ \sum_{k=0}^{(p-1)/2}(-1)^k(4k+1)A_k\equiv p(-1)^{(p-1)/2}\pmod{p^3} \quad\text{and}\quad…

In this paper we obtain some congruences involving central binomial coefficients and Lucas sequences. For example, we show that if p>5 is a prime then $\sum_{k=0}^{p-1}F_k*binom(2k,k)/12^k$ is congruent to 0,1,-1 modulo p according as p=1,4…

We give elementary proofs for the Apagodu-Zeilberger-Stanton-Amdeberhan-Tauraso congruences $$\sum\limits_{n=0}^{p-1}\dbinom{2n}{n} \equiv\eta_{p}\mod p^{2},$$ $$\sum\limits_{n=0}^{rp-1}\dbinom{2n}{n}…

Let $\{A'_n\}$ be the Ap\'ery numbers given by $A'_n=\sum_{k=0}^n\binom nk^2\binom{n+k}k.$ For any prime $p\equiv 3\pmod 4$ we show that $A'_{\frac{p-1}2}\equiv \frac{p^2}3\binom{\frac{p-3}2}{\frac{p-3}4}^{-2}\pmod {p^3}$. Let $\{t_n\}$ be…

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…

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…

Let $p$ be an odd prime, Jianqiang Zhao has established a curious congruence $$ \sum_{i+j+k=p \atop i,j,k > 0} \frac{1}{ijk} \equiv -2B_{p-3}\pmod p , $$ where $B_{n}$ denotes the $n-$th Bernoulli numbers. In this paper, we will generalize…

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…

In this paper, we partly prove a supercongruence conjectured by Z.-W. Sun in 2013. Let $p$ be an odd prime and let $a\in\mathbb{Z}^{+}$. Then if $p\equiv1\pmod3$, we have \begin{align*}…

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…

In this paper we study some sophisticated supercongruences involving dual sequences. For $n=0,1,2,\ldots$ define $$d_n(x)=\sum_{k=0}^n\binom nk\binom xk2^k$$ and $$s_n(x)=\sum_{k=0}^n\binom nk\binom xk\binom{x+k}k=\sum_{k=0}^n\binom…

The Ap\'ery polynomials are given by $$A_n(x)=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2x^k\ \ (n=0,1,2,\ldots).$$ (Those $A_n=A_n(1)$ are Ap\'ery numbers.) Let $p$ be an odd prime. We show that…

Suppose that $p$ is an odd prime and $m$ is an integer not divisible by $p$. Sun and Tauraso [Adv. in Appl. Math., 45(2010), 125--148] gave $\sum_{k=0}^{n-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=0}^{n-1}\binom{2k}{k+d}/(km^k)$ modulo $p$ for…

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 mainly prove two congruence conjecture of Z.-W. Sun. Let $p\equiv3\pmod 4$ be a prime. Then $$\sum_{k=0}^{p-1}\frac{\binom{2k}k^2}{8^k}\equiv-\sum_{k=0}^{p-1}\frac{\binom{2k}k^2}{(-16)^k}\pmod{p^3}.$$ And for any odd prime…

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…

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…