Related papers: On a kind of generilized multi-harmonic sum

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

Let $p$ be a prime with $p>3$, and let $a,b$ be two rational $p-$integers. In this paper we present general congruences for $\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac p{k+b}\pmod {p^2}$. For $n=0,1,2,\ldots$ let $D_n$ and $b_n$ be Domb…

We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd}…

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))…

Let $p>3$ be a prime. In this paper, we obtain the congruences for $$\sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^3}{(-8)^k},\ \sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^2\binom{3k}k}{(-192)^k},\…

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 $$…

The Delannoy numbers and Schr\"oder numbers are given by \begin{align*} D_n=\sum_{k=0}^n{n\choose k}{n+k\choose k}\quad \text{and}\quad S_n=\sum_{k=0}^n{n\choose k}{n+k\choose k}\frac{1}{k+1}, \end{align*} respectively. Let $p>3$ be a…

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…

Let p be any prime, and let a and n be nonnegative integers. Let $r\in Z$ and $f(x)\in Z[x]$. We establish the congruence $$p^{\deg f}\sum_{k=r(mod p^a)}\binom{n}{k}(-1)^k f((k-r)/p^a) =0 (mod p^{\sum_{i=a}^{\infty}[n/p^i]})$$ (motivated by…

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 n be a positive odd integer and let p>n+1 be a prime. We mainly derive the following congruence: $$\sum_{0<i_1<...<i_n<p}(i_1/3)(-1)^{i_1}/(i_1...i_n)=0 (mod p).$$

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*}…

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,…

In this paper we prove three results conjectured by Z.-W. Sun. Let $p$ be an odd prime and let $h\in \mathbb{Z}$ with $2h-1\equiv0\pmod{p^{}}$. For $a\in\mathbb{Z}^{+}$ and $p^a>3$, we show that \begin{align}\notag…

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…

Let $p$ be an odd prime, and let $m$ be an integer with $p\nmid m$. In this paper show that $$\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak\binom{-1-a}k}{m^k} \equiv 0\pmod p \quad\hbox{implies}\quad\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak…

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 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 $p>3$ 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 show that for any prime prime $p\not=2$ $$\sum_{k=1}^{p-1} {(-1)^k\over k}{-{1\over 2} \choose k} \equiv -\sum_{k=1}^{(p-1)/2}{1\over k} \pmod{p^3}$$ by expressing the l.h.s. as a combination of alternating multiple harmonic sums.