相关论文: An Extension of a Congruence by Kohnen

Let $p\ge 5$ be a prime, and let $q_p(2):=(2^{p-1}-1)/p$ be the Fermat quotient of $p$ to base 2. The following curious congruence was conjectured by L. Skula and proved by A. Granville $$ q_p(2)^2\equiv…

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,…

We prove some polynomial identities from which we deduce congruences modulo $p^2$ for the Fermat quotient $\frac{2^p-2}{p}$ for any odd prime $p$ (Proposition 1 and Theorem 1). These congruences are simpler than the one obtained by…

Given an odd prime p, we prove a congruence on the p-residue of the Fermat quotient q_p(a) in base a with 0<a<p, which arises from a generalization of the Voronoi congruences and from some other congruences on sums and weighted sums of…

We derive $p$-adic expansions for the generalized Harmonic numbers $H^{(j)}_{p-1}$ and $H^{(j)}_{\frac{p-1}{2}}$ involving the Bernoulli numbers $B_j$ and the the base-2 Fermat quotient $q_p$. While most of our results are not new, we…

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…

We prove several supercongruences involving the harmonic number of order two $H_n^{(2)}:=\sum_{k=1}^n1/k^2$. For example, if $p>5$ is prime and $\alpha$ is $p$-integral, then we can completely determine $$…

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…

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…

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/n$ be the $n$th harmonic number. In this note we prove that for any prime $p\ge 7$, $$ \sum_{k=1}^{p-1}\frac{H_k}{k^2}\equiv \sum_{k=1}^{p-1}\frac{H_k^2}{k}…

Recently the first author proved a congruence proposed in 2006 by Adamchuk: $\sum_{k=1}^{\lfloor\frac{2p}{3}\rfloor}\binom{2k}{k}\equiv 0\pmod{p^2}$ for any prime $p=1 \pmod{3}$. In this paper, we provide more examples (with proofs) of…

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…

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 $p>3$, $$…

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/k$ be the $n$th harmonic number. In this note we prove that for any prime $p\ge 7$, $$ \sum_{k=1}^{p-1}\frac{H_k^2}{k^2} \equiv4/5pB_{p-5}\pmod{p^2}, $$ which confirms the conjecture…

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…

The Franel numbers are defined by $ f_n=\sum_{k=0}^n {n\choose k}^3. $ Motivated by the recent work of Z.-W. Sun on Franel numbers, we prove that \begin{align*} \sum_{k=0}^{n-1}(3k+1)(-16)^{n-k-1} {2k\choose k} f_k &\equiv…

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 $n$)…

In this paper, we mainly prove the following conjectures of Z.-W. Sun \cite{S13}: Let $p>2$ be a prime. If $p=x^2+3y^2$ with $x,y\in\mathbb{Z}$ and $x\equiv1\pmod 3$, then $$x\equiv\frac14\sum_{k=0}^{p-1}(3k+4)\frac{f_k}…

In this paper we prove that for any prime $p\ge 11$ holds $$ {2p-1\choose p-1}\equiv 1 -2p \sum_{k=1}^{p-1}\frac{1}{k} +4p^2\sum_{1\le i<j\le p-1}\frac{1}{ij}\pmod{p^7}. $$ This is a generalization of the famous Wolstenholme's theorem which…

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,…