Related papers: New series involving binomial coefficients (II)

Let p be an odd prime and let a be a positive integer. In this paper we investigate the sum $\sum_{k=0}^{p^a-1}\binom{hp^a-1}{k}\binom{2k}{k}/m^k$ mod p^2, where h,m are p-adic integers with m\not=0 (mod p). For example, we show that if…

We present various identities involving the classical Bernoulli and Euler polynomials. Among others, we prove that $$ \sum_{k=0}^{[n/4]}(-1)^k {n\choose 4k}\frac{B_{n-4k}(z) }{2^{6k}} =\frac{1}{2^{n+1}}\sum_{k=0}^{n} (-1)^k…

We consider a special class of binomial sums involving harmonic numbers and we prove three identities by using the elementary method of the partial fraction decomposition. Some applications to infinite series and congruences are given.

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 derive modular parametrizations for certain infinite series whose summands involve central binomial coefficients and higher-order harmonic numbers. When the rates of convergence are certain rational numbers, modularity allows us to…

Let $\{a_{1}, a_{2},\ldots, a_{n},\ldots\}$ be a sequence of complex numbers which has at most polynomial growth and satisfies an extra assumption. In this paper, inspired by a recent work of Sasane, we give an explanation of the sum…

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. For any $p$-adic integer $a$, we determine $$\sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k,\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k^{(2)},\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}k\frac{H_k^{(2)}}{2k+1}$$ modulo…

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…

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…

We show how to find series expansions for $\pi$ of the form $\pi=\sum_{n=0}^\infty {S(n)}\big/{\binom{mn}{pn}a^n}$, where S(n) is some polynomial in $n$ (depending on $m,p,a$). We prove that there exist such expansions for $m=8k$, $p=4k$,…

We present several congruences modulo a power of prime $p$ concerning sums of the following type $\sum_{k=1}^{p-1}{m^k\over k^r}{2k\choose k}^{-1}$ which reveal some interesting connections with the analogous infinite series.

Let $p$ be an odd prime and let $a,m$ be integers with $a>0$ and $m \not\equiv0\pmod p$. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ mod $p^2$ for $d=0,1$; for example,…

In this paper we study recurrences concerning the combinatorial sum $[n,r]_m=\sum_{k\equiv r (mod m)}\binom {n}{k}$ and the alternate sum $\sum_{k\equiv r (mod m)}(-1)^{(k-r)/m}\binom{n}{k}$, where m>0, $n\ge 0$ and r are integers. For…

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…

Via symbolic summation method, we establish the following series for $\pi^2$: \begin{align*} \sum_{k=1}^\infty \frac{H_k-2H_{2k}}{(-3)^k k} = \frac{\pi^2}{18}, \end{align*} where $H_k=\sum_{j=1}^k 1/j$. We also derive a $p$-adic congruence…

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 work we prove a new combinatorial identity and applying it we establish many finite harmonic sum identities. Among many others, we prove that \begin{equation*}…

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…

We give a characterization of all pairs $(k,n)$ of positive integers for which the ratio $$ \frac{1^k-2^k+3^k-\dots+(-1)^{n+1} n^k}{1^k-2^k+3^k-\dots+(-1)^{n}(n-1)^k} $$ of two consecutive alternating power sums is an integer.