相关论文: On two conjectural supercongruences of Apagodu and…

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 2014, Wang and Cai established the following harmonic congruence for any odd prime $p$ and positive integer $r$, \begin{equation*} Z(p^{r})\equiv-2p^{r-1}B_{p-3} ~(\bmod ~ p^{r}), \end{equation*} where $…

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 supercongruences by the Wilf-Zeilberger method. One of them is, for any prime $p>3$, \begin{align*} \sum_{n=0}^{(p-1)/2}\frac{3n+1}{(-8)^n}\binom{2n}n^3\equiv…

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…

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…

Let $p$ be an odd prime and let $n$ be a positive integer. For any positive integer $\alpha$ and $m\in\{1,2,3\}$, we have \begin{align*}…

In this paper we establish some congruences involving the Ap\'ery numbers $\beta_{n}=\sum_{k=0}^{n}\binom{n}{k}^2\binom{n+k}{k}$ $(n=0,1,2,\ldots)$. For example, we show that $$\sum_{k=0}^{n-1}(11k^2+13k+4)\beta_k\equiv0\pmod{2n^2}$$ for…

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…

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 $p>3$, $$…

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…

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 $2012$, Guillera and Zudilin established the following two supercongruences involving truncated Ramanujan-type series: for any odd prime $p>2$, \begin{align*}…

For $A,B\in\mathbb Z$, the Lucas sequence $u_n(A,B)\ (n=0,1,2,\ldots)$ are defined by $u_0(A,B)=0$, $u_1(A,B)=1$, and $u_{n+1}(A,B) = Au_n(A,B)-Bu_{n-1}(A,B)$ $(n=1,2,3,\ldots).$ For any odd prime $p$ and positive integer $n$, we establish…

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,…

We give elementary proofs for the Apagodu-Zeilberger-Stanton-Amdeberhan-Tauraso congruences $$\sum\limits_{n=0}^{p-1}\dbinom{2n}{n} \equiv\eta_{p}\mod p^{2},$$ $$\sum\limits_{n=0}^{rp-1}\dbinom{2n}{n}…

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.

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…

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

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…