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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…

Number Theory · Mathematics 2018-04-10 Romeo Mestrovic

We give a family of congruences for the binomial coefficients ${kp-1\choose p-1}$ in terms of multiple harmonic sums, a generalization of the harmonic numbers. Each congruence in this family (which depends on an additional parameter $n$)…

Number Theory · Mathematics 2018-10-16 Julian Rosen

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 $p>3$, $$…

Number Theory · Mathematics 2020-06-30 Chen Wang

Let p be a prime and let a be a positive integer. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=1}^{p-1}\binom{2k}{k+d}/(km^{k-1})$ modulo $p$ for all d=0,...,p^a, where m is any integer not divisible by p.…

Number Theory · Mathematics 2010-04-02 Zhi-Wei Sun , Roberto Tauraso

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

Number Theory · Mathematics 2025-12-03 Jiaqi Wang , Rong Ma

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/k$ be the $n$th harmonic number. Z. W. Sun conjectured that for any prime $p\ge 5$, $$ \sum_{k=1}^{p-1}\frac{H_k}{k\cdot 2^k} \equiv7/24pB_{p-3}\pmod{p^2}. $$ This conjecture is recently…

Number Theory · Mathematics 2018-04-10 Romeo Mestrovic

In this paper, we prove that for any odd prime $p$ and for any $p$-integer $\alpha $,we have $ \binom{\alpha p-1}{p-1}\equiv 1-\alpha (\alpha -1)(\alpha ^{2}-\alpha -1)p\sum_{k=1}^{p-1}\frac{1}{k}+\alpha ^{2} (\alpha-1)^{2}p^{2}\sum_{1\leq…

Combinatorics · Mathematics 2016-07-05 Farid Bencherif , Rachid Boumahdi

Let $p$ be an odd prime. In 2008 E. Mortenson proved van Hamme's following conjecture: $$\sum_{k=0}^{(p-1)/2}(4k+1)\binom{-1/2}k^3\equiv (-1)^{(p-1)/2}p\pmod{p^3}.$$ In this paper we show further that…

Number Theory · Mathematics 2014-02-19 Zhi-Wei Sun

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.

Number Theory · Mathematics 2021-09-22 Guo-Shuai Mao , Zhi-Wei Sun

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,…

Number Theory · Mathematics 2018-01-08 Nicholas J. Newsome , Maria S. Nogin , Adnan H. Sabuwala

In this paper we deduce some new supercongruences modulo powers of a prime $p>3$. Let $d\in\{0,1,\ldots,(p-1)/2\}$. We show that $$\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{8^k}\equiv 0\ (\mbox{mod}\ p)\ \ \ \mbox{if}\ d\equiv…

Number Theory · Mathematics 2013-10-31 Zhi-Wei Sun

The polynomials $d_n(x)$ are defined by \begin{align*} d_n(x) &= \sum_{k=0}^n{n\choose k}{x\choose k}2^k. \end{align*} We prove that, for any prime $p$, the following congruences hold modulo $p$: \begin{align*}…

Number Theory · Mathematics 2016-04-19 Song Guo , Victor J. W. Guo

In this paper we prove three results conjectured by Z.-W. Sun. Let $p$ be an odd prime and let $h\in \mathbb{Z}$ with $2h-1\equiv0\pmod{p^{}}$. For $a\in\mathbb{Z}^{+}$ and $p^a>3$, we show that \begin{align}\notag…

Combinatorics · Mathematics 2019-11-04 Yong Zhang

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…

Number Theory · Mathematics 2012-10-09 Khodabakhsh Hessami Pilehrood , Tatiana Hessami Pilehrood , Roberto Tauraso

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…

Number Theory · Mathematics 2026-01-26 Guo-Shuai Mao

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}}…

Number Theory · Mathematics 2018-04-06 Jianqiang Zhao

Let $p>3$ be a prime and let $a$ be a positive integer. We show that if $p\equiv1\pmod 4$ or $a>1$ then $$\sum_{k=0}^{\lfloor\frac34p^a\rfloor}\frac{\binom{2k}k^2}{16^k}\equiv\l(\frac{-1}{p^a}\r)\pmod{p^3}$$ with $(-)$ the Jacobi symbol,…

Number Theory · Mathematics 2018-09-25 Guo-Shuai Mao , Zhi-Wei Sun

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…

Number Theory · Mathematics 2024-01-11 Zhi-Wei Sun

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…

Number Theory · Mathematics 2023-04-11 Guo-Shuai Mao

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}\ \ \…

Number Theory · Mathematics 2015-02-27 Zhi-Wei Sun
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