Related papers: Some Congruences Involving Binomial Coefficient an…

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…

In this paper we establish some new congruences involving central binomial coefficients as well as Catalan numbers. Let $p$ be a prime and let $a$ be any positive integer. We determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}$ mod $p^2$ for…

In this paper, we evaluate some series of the form $$\sum_{k=1}^\infty\frac{ak^2+bk+c}{k(3k-1)(3k-2)m^k\binom{4k}k}.$$ For example, we prove that $$\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^{k}}{k(3k-1)(3k-2)\binom{4k}k}=\frac{3}2\pi$$ and…

In this paper, we establish the following two identities involving the Gamma function and Bernoulli polynomials, namely $$ \sum_{k\leq x}\frac{1}{k^s} \sum_{j=1}^{k^s}\log\Gamma\left(\frac{j}{k^s}\right) \sum_{\substack{d|k \\…

Let $p>5$ be a prime. Motivated by the known formulae $\sum_{k=1}^\infty(-1)^k/(k^3\binom{2k}{k})=-2\zeta(3)/5$ and $\sum_{k=0}^\infty \binom{2k}{k}^2/((2k+1)16^k)=4G/\pi$$ (where $G=\sum_{k=0}^\infty(-1)^k/(2k+1)^2$ is the Catalan…

Let $p>3$ be a prime, and let $q_p(2)=(2^{p-1}-1)/p$ be the Fermat quotient of $p$ to base 2. Recently, Z. H. Sun proved that \sum_{k=1}^{p-1}\frac{1}{k\cdot 2^k}\equiv q_p(2)-\frac{p}{2}q_p(2)^2 \pmod{p^2} which is a generalization of a…

In this note we associate a sequence of non-negative integers to any convergent series of positive real numbers and study this sequence for the series $\sum_{n \geq 1} n^{-k}$ where $k$ is an integer $\geq 2$.

Let $ p \ge 5 $ be a prime and let $ b, c \in \mathbb{Z} $. Denote by $ T_k(b,c) $ the generalized central trinomial coefficient, i.e., the coefficient of $ x^k $ in $ (x^2 + bx + c)^k $. In this paper, we establish congruences modulo $ p^3…

Let $p$ be a prime and ${\mathcal{P}_{p}}$ the set of positive integers which are prime to $p$. We establish the following interesting congruence \[\sum\limits_{\begin{smallmatrix} i+j+k={{p}^{r}} i,j,k\in {\mathcal{P}_{p}}…

We study divisibility properties of certain sums and alternating sums involving binomial coefficients and powers of integers. For example, we prove that for all positive integers $n_1,..., n_m$, $n_{m+1}=n_1$, and any nonnegative integer…

Let $p$ be a prime and let $x$ be a $p$-adic integer. We provide two supercongruences for truncated series of the form $$\sum_{k=1}^{p-1} \frac{(x)_k}{(1)_k}\cdot \frac{1}{k}\sum_{1\le j_1\le\cdots\le j_r\le k}\frac{1}{j_1^{}\cdots…

Let $p$ be an odd prime, and let $a$ be a rational $p$-adic integer with $a\not\equiv 0\pmod p$. In this paper, using WZ method we establish the congruences for $\sum_{k=0}^{p-1} \binom ak^2(-1)^k(1-\frac 2ak)$ modulo $p^2$ and…

It is significant to study congruences involving multiple harmonic sums. Let $p$ be an odd prime, in recent years, the following curious congruence $$\sum_{\substack{i+j+k=p \\ i, j, k>0}} \frac{1}{i j k} \equiv-2 B_{p-3}\pmod p$$ has been…

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…

We explore new types of binomial sums with Fibonacci and Lucas numbers. The binomial coefficients under consideration are $\frac{n}{n+k}\binom{n+k}{n-k}$ and $\frac{k}{n+k}\binom{n+k}{n-k}$. The identities are derived by relating the…

In this paper we obtain some congruences involving central binomial coefficients and Lucas sequences. For example, we show that if p>5 is a prime then $\sum_{k=0}^{p-1}F_k*binom(2k,k)/12^k$ is congruent to 0,1,-1 modulo p according as p=1,4…

In this paper we mainly employ the Zeilberger algorithm to study congruences for sums of terms involving products of three binomial coefficients. Let $p>3$ be a prime. We prove that…

With help of $q$-congruence, we prove the divisibility of some binomial sums. For example, for any integers $\rho,n\geq 2$, $$\sum_{k=0}^{n-1}(4k+1) \binom{2k}{k}^\rho \cdot (-4)^{\rho(n-1-k)} \equiv 0\pmod{2^{\rho-2}n\binom{2n}{n}}.$$

For a pair of positive integers $n,k$ with $n\geq 2$, in this paper we prove that $$ \sum_{r=1}^k\sum_{|\bf\alpha|=k}{k\choose\bf\alpha} \zeta(n\bf\alpha)=\zeta(n)^k =\sum^k_{r=1}\sum_{|\bf\alpha|=k}…

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…