Related papers: Proof of two supercongruences by the Wilf-Zeilberg…

In this paper, we prove two supercongruences conjectured by Z.-W. Sun via the Wilf-Zeilberger method. One of them is, for any prime $p>3$, \begin{align*} \sum_{n=0}^{(p-1)/2}\frac{6n+1}{(-512)^n}\binom{2n}n^3&\equiv…

In this paper, we prove two supercongruences conjectured by Z.-W. Sun via the Wilf-Zeilberger method. One of them is, for any prime $p>3$, \begin{align*} \sum_{n=0}^{p-1}\frac{6n+1}{256^n}\binom{2n}n^3&\equiv…

In this paper, we employ the Wilf-Zeilberger (WZ) method to prove a supercongruence conjecture posed by Z.-W. Sun: for any prime $p$, \begin{align*} \sum_{k=0}^{\frac{p-3}{2}}\frac{92k^2+61k+9}{(2k+1)64^k}{2k \choose k}{3k \choose k}{4k…

In this paper, we mainly prove two conjectural supercongruences of Sun by using the following identity $$ \sum_{k=0}^n\binom{2k}{k}^2\binom{2n-2k}{n-k}^2=16^n\sum_{k=0}^n\frac{\binom{n+k}{k}\binom{n}{k}\binom{2k}{k}^2}{(-16)^k} $$ which…

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…

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 $p$ 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 $$…

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…

In this paper, we prove some supercongruences via the Wilf-Zeilberger method. For instance, for any odd prime $p$ and positive integer $r$ and $\delta\in\{1,2\}$, we have \begin{align*} \sum_{n=0}^{(p^r-1)/\delta}…

We establish some supercongruences related to a supercongruence of Van Hamme, such as \begin{align*} \sum_{k=0}^{(p+1)/2} (-1)^k (4k-1)\frac{(-\frac{1}{2})_k^3}{k!^3} &\equiv p(-1)^{(p+1)/2}+p^3(2-E_{p-3})\pmod{p^{4}},\\…

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…

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

Using Zeilberger's algorithm, we here give a proof of the supercongruence $$ \sum_{n=0}^{\frac{p^r-3}{4}}(8n+1)\frac{\left(\frac{1}{4}\right)_n^4}{(1)_n^4}\equiv -p^3…

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…

The numbers $R_n$ and $W_n$ are defined as \begin{align*} R_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{1}{2k-1},\ \text{and}\ W_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{3}{2k-3}. \end{align*} We prove that, for any positive…

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 mainly prove the following conjecture of Z.-H. Sun cite{SH20}: Let $p>3$ be a prime. Then $$\sum_{k=0}^{p-1}\binom{2k}k\frac{3k+1}{(-16)^k}f_k\equiv(-1)^{(p-1)/2}p+p^3E_{p-3}\pmod{p^4},$$ where…

Let $M_n$ and $T_n$ denote the $n$th Motzkin number and the $n$th central trinomial coefficient respectively. We prove that for any prime $p\ge 5$, \begin{align*} &\sum_{k=0}^{p-1}M_k^2\equiv…

We utilize the Wilf-Zeilberger (WZ) method to establish congruences related to truncated Ramanujan-type series. By constructing hypergeometric terms $f(k, a, b, \ldots)$ with Gosper-summable differences and selecting appropriate parameters,…

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…