相关论文: A Generalized Supercongruence of Z.-W. Sun

In this paper, we prove two conjectures of Z.-W. Sun: $$2n\binom{2n}n\big|\sum_{k=0}^{n-1}(3k+1)\binom{2k}k^3{16}^{n-1-k}\ \mbox{for}\ \mbox{all}\ n=2,3,\cdots,$$ and $$\sum_{k=0}^{(p-1)/2}\frac{3k+1}{16^k}\binom{2k}{k}^3\equiv…

Let $E_n$ be the $n$-th Euler number and $(a)_n=a(a+1)\cdots (a+n-1)$ the rising factorial. Let $p>3$ be a prime. In 2012, Sun proved the that $$ \sum^{(p-1)/2}_{k=0}(-1)^k(4k+1)\frac{(\frac{1}{2})_k^3}{k!^3} \equiv…

Let $p$ be an odd prime and let $x$ be a $p$-adic integer. In this paper, we establish supercongruences for $$ \sum_{k=0}^{p-1}\frac{\binom{x}{k}\binom{x+k}{k}(-4)^k}{(dk+1)\binom{2k}{k}}\pmod{p^2} $$ and $$…

Recently, Z.-W. Sun made the following conjecture: for any odd prime $p$ and odd integer $m$, $$ \frac{1}{m^2{m-1\choose (m-1)/2}}\Bigg(\sum_{k=0}^{(pm-1)/2}\frac{{2k\choose k}}{8^k}…

For a non-negative integer $m$, let $S(m)$ denote the sum given by $$S(m):=\sum_{n=0}^{m}\frac{(-1)^n(8n+1)}{n!^3}\left(\frac{1}{4}\right)_n^3.$$ Using the powerful WZ-method, for a prime $p\equiv 3$ $($mod $4)$ and an odd integer $r>1$, we…

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

In this paper, we mainly establish a congruence for a sum involving Ap\'{e}ry numbers, which was conjectured by Z.-W. Sun. Namely, for any prime $p>3$ and positive odd integer $m$, we prove that there is a $p$-adic integer $c_m$ only…

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…

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 $$…

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,…

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…

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…

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 $p>3$ 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…

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 $$…

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…

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: $$…

Let the numbers $\alpha_n,\beta_n$ and $\gamma_n$ denote \begin{align*} \alpha_n=\sum_{k=0}^{n-1}{2k\choose k},\quad \beta_n=\sum_{k=0}^{n-1}{2k\choose k}\frac{1}{k+1}\quad\text{and}\quad \gamma_n=\sum_{k=0}^{n-1}{2k\choose…

In this paper, we study some supercongruences involving the sequence $$ t_n(x)=\sum_{k=0}^n\binom{n}{k}\binom{x}{k}\binom{x+k}{k}2^k $$ and solve some open problems. For any odd prime $p$ and $p$-adic integer $x$, we determine…

Let $p>3$ be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{2k}{k}/2^k=(-1)^{(p-1)/2}-p^2E_{p-3} (mod p^3),$$ $$\sum_{k=1}^{(p-1)/2}\binom{2k}{k}/k=(-1)^{(p+1)/2}8/3*pE_{p-3} (mod p^2),$$…