相关论文: On a kind of generalized multi-harmonic sums

We derive $p$-adic expansions for the generalized Harmonic numbers $H^{(j)}_{p-1}$ and $H^{(j)}_{\frac{p-1}{2}}$ involving the Bernoulli numbers $B_j$ and the the base-2 Fermat quotient $q_p$. While most of our results are not new, we…

In this paper we prove that for any prime $p\ge 11$ holds $$ {2p-1\choose p-1}\equiv 1 -2p \sum_{k=1}^{p-1}\frac{1}{k} +4p^2\sum_{1\le i<j\le p-1}\frac{1}{ij}\pmod{p^7}. $$ This is a generalization of the famous Wolstenholme's theorem which…

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…

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

The harmonic numbers $H_n=\sum_{0<k\le n}1/k\ (n=0,1,2,\ldots)$ play important roles in mathematics. Let $p>3$ be a prime. With helps of some combinatorial identities, we establish the following two new congruences:…

Let $p>3$ be a prime, and let $a$ be a rational p-adic integer with $a\not\equiv 0\pmod p$. In this paper we establish congruences for $$\sum_{k=1}^{(p-1)/2}\frac{\binom ak\binom{-1-a}k}k, \quad\sum_{k=0}^{(p-1)/2}k\binom ak\binom{-1-a}k…

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…

Let $m$, $r$ and $n$ be positive integers. We denote by ${\bf k}\vdash n$ any tuple of odd positive integers ${\bf k}=(k_1,\dots,k_t)$ such that $k_1+\dots+k_t=n$ and $k_j\ge 3$ for all $j$. In this paper we prove that for every…

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…

The following congruence for power sums, $S_n(p)$, is well known and has many applications: $1^n+2^n +\dots +p^n \equiv\begin{cases} -1 \text{ mod } p, & \text{ if } \ p-1 \ | \ n; 0 \text{ mod } p, & \text{ if } \ p-1 \ \not| \ n,…

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. Define the Gaussian power sum \[ G_n(p)=\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}(a+bi)^n\in\mathbb Z[i]. \] We determine $G_p(p)$ modulo high powers of $p$: if $p\equiv 1\pmod 4$ then $$G_p(p)\equiv p^2(1+i)\pmod{p^3},$$…

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…

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 $p>3$ 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 $p>3$ 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 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.

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 $$…

We produce congruences modulo a prime $p>3$ for sums $\sum_k\binom{3k}{k}x^k$ over ranges $0\le k<q$ and $0\le k<q/3$, where $q$ is a power of $p$. Here $x$ equals either $c^2/(1-c)^3$, or $4s^2/\bigl(27(s^2-1)\bigr)$, where $c$ and $s$ are…

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.