Related papers: Note on super congruences modulo $p^2$

Let $\{f_n\}$ be the Franel numbers given by $f_n=\sum_{k=0}^n\binom nk^3$, and let $p>5$ be a prime. In this paper we mainly determine $\sum_{k=0}^{p-1} \binom{2k}k\frac{f_k}{m^k}\pmod p$ for $m=5,-16,16,32,-49,50,96$. Let…

For $n=0,1,2,\ldots$ let $W_n=\sum_{k=0}^{[n/3]}\binom{2k}k \binom{3k}k\binom n{3k}(-3)^{n-3k}$, where $[x]$ is the greatest integer not exceeding $x$. Then $\{W_n\}$ is an Ap\'ery-like sequence. In this paper we deduce many congruences…

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}\ \ \…

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that $\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ and $\nu_p(\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{\binom{2k}k}{m^k})$ are at…

In this paper we establish some sophisticated congruences involving central binomial coefficients and Fibonacci numbers. For example, we show that if $p\not=2,5$ is a prime then $$\sum_{k=0}^{p-1}F_{2k}\binom{2k}{k}=(-1)^{[p/5]}(1-(p/5))…

Suppose that $p$ is an odd prime and $m$ is an integer not divisible by $p$. Sun and Tauraso [Adv. in Appl. Math., 45(2010), 125--148] gave $\sum_{k=0}^{n-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=0}^{n-1}\binom{2k}{k+d}/(km^k)$ modulo $p$ for…

Let $p$ be a prime and let $a$ be a positive integer. In this paper we investigate $\sum_{k=0}^{p^a-1}\binom[(h+1)k,k+d]/m^k$ modulo a prime $p$, where $d$ and $m$ are integers with $-h<d<=p^a$ and $m\not=0 (mod p)$. We also study…

We prove that $\sum_{k=0}^{q-1}\binom{2k}{k}\equiv q^2\pmod{3q^2}$ if q>1 is a power of 3, as recently conjectured by Z.W. Sun and R. Tauraso. Our more precise result actually implies that the value of $(1/q^2)\sum_{k=0}^{q-1}\binom{2k}{k}$…

We show that if p is an odd prime then $$\sum_{k=0}^{p-1}E_kE_{p-1-k}=1 (mod p)$$ and $$\sum_{k=0}^{p-3}E_kE_{p-3-k}=(-1)^{(p-1)/2}2E_{p-3} (mod p),$$ where E_0,E_1,E_2,... are Euler numbers. Moreover, we prove that for any positive integer…

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}…

Let $p>3$ be a prime, and let $m$ be an integer with $p\nmid m$. In the paper, by using the work of Ishii and Deuring's theorem for elliptic curves with complex multiplication we solve some conjectures of Zhi-Wei Sun concerning…

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…

For any positive integers $m$ and $\alpha$, we prove that $$\sum_{k=0}^{n-1}\epsilon^k(2k+1)A_k^{(\alpha)}(x)^m\equiv0\pmod{n}, $$ where $\epsilon\in\{1,-1\}$ and $$…

In this paper, we partly prove a supercongruence conjectured by Z.-W. Sun in 2013. Let $p$ be an odd prime and let $a\in\mathbb{Z}^{+}$. Then if $p\equiv1\pmod3$, we have \begin{align*}…

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}…

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 present some congruences modulo $p^{6-d}$ for sums of the type $\sum_{k=0}^{(p-3)/2}x^k{2k\choose k}/(2k+1)^d$, for $d=1,2,3$ where $p>5$ is a prime.

We mainly introduce two new kinds of numbers given by $$R_n=\sum_{k=0}^n\binom nk\binom{n+k}k\frac1{2k-1}\quad\ (n=0,1,2,...)$$ and $$S_n=\sum_{k=0}^n\binom nk^2\binom{2k}k(2k+1)\quad\ (n=0,1,2,...).$$ We find that such numbers have many…

Using an identity arising in the known Banach probability problem on matchboxes, we prove an unexpected congruence for odd prime $p:$ for $1\leq k\leq \frac{p-1}{2},\enskip \sum_{i=1}^{p-2k-1}2^{i-1}\binom{k-1+i}{k}\equiv 0\pmod p.$

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.