Related papers: Arithmetic theory of harmonic numbers (II)

We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd}…

In this paper we mainly employ the Zeilberger algorithm to study congruences for sums of terms involving products of three binomial coefficients. Let $p>3$ be a prime. We prove that…

In this paper, we mainly prove the following conjectures of Z.-H. Sun \cite{SH2}: Let $p>3$ be a prime. If $p\equiv1\pmod3$ and $p=x^2+3y^2$, then we have $$…

We mainly introduce two new kinds of numbers given by $$R_n=\sum_{k=0}^n\binom nk\binom{n+k}k\frac1{2k-1}\quad\ (n=0,1,2,...)$$ and $$S_n=\sum_{k=0}^n\binom nk^2\binom{2k}k(2k+1)\quad\ (n=0,1,2,...).$$ We find that such numbers have many…

In this paper, we confirm some congruences conjectured by V.J.W. Guo and M.J. Schlosser recently. For example, we show that for primes $p>3$, $$…

Let $\{f_n\}$ be the Franel numbers given by $f_n=\sum_{k=0}^n\binom nk^3$, and let $p>5$ be a prime. In this paper we mainly determine $\sum_{k=0}^{p-1} \binom{2k}k\frac{f_k}{m^k}\pmod p$ for $m=5,-16,16,32,-49,50,96$. Let…

It is known that $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)4^k)=\pi/2$ and $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)16^k)=\pi/3$. In this paper we obtain their p-adic analogues such as…

In 2014, Wang and Cai established the following harmonic congruence for any odd prime $p$ and positive integer $r$, \begin{equation*} \sum\limits_{i+j+k=p^{r}\atop{i,j,k\in \mathcal{P}_{p}}}\frac{1}{ijk}\equiv-2p^{r-1}B_{p-3} ~(\bmod ~…

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…

In this paper, we prove several supercongruences conjectured by Z.-W. Sun ten years ago via certain strange hypergeometric identities. For example, for any prime $p>3$, we show that…

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…

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 $p>5$ be a prime. We prove congruences modulo $p^{3-d}$ for sums of the general form $\sum_{k=0}^{(p-3)/2}\binom{2k}{k}t^k/(2k+1)^{d+1}$ and $\sum_{k=1}^{(p-1)/2}\binom{2k}{k}t^k/k^d$ with $d=0,1$. We also consider the special case…

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…

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…

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…

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…

Recently, Mneimneh proved the remarkable identity \begin{align*} \sum_{k=0}^n H_k\binom{n}{k} p^k(1-p)^{n-k}=\sum_{i=1}^n \frac{1-(1-p)^i}{i}\quad (p\in [0,1]) \end{align*} as the main result of a 2023 \emph{Discrete Mathematics} paper,…

Let $p>3$ be a prime, and let $a$ be a rational $p$-adic integer, using WZ method we establish the congruences modulo $p^3$ for $$\sum_{k=0}^{p-1} \binom ak\binom{-1-a}k\binom{2k}k\frac {w(k)}{4^k},$$ where $$w(k)=1,\frac 1{k+1},\frac…

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…