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We give a direct combinatorial proof of a famous identity, $$ \sum_{i+j=n} m{2i}{i} \binom{2j}{j} = 4^n $$ by actually counting pairs of $k$-subsets of $2k$-sets. Then we discuss two different generalizations of the identity, and end the…

组合数学 · 数学 2016-11-22 Rui Duarte , António Guedes de Oliveira

We show that double sums of the form $$ \sum_{i,j=-n} ^{n} |i^sj^t(i^k-j^k)^\beta| \binom {2n} {n+i} \binom {2n} {n+j} $$ can always be expressed in terms of a linear combination of just four functions, namely $\binom {4n}{2n}$, ${\binom…

组合数学 · 数学 2020-12-07 Christian Krattenthaler , Carsten Schneider

We explain how the identity $$\sum_{i+j=n}\binom{2i}{i}\binom{2j}{j}\;=\;\displaystyle4^n$$ is an easy consequence of the inclusion-exclusion principle.

组合数学 · 数学 2013-07-26 Rui Duarte , António Guedes de Oliveira

We will show in this text that, for all non-negative integers $n$ and $l$, the following equality is verified: \[\sum_{i=0}^{l} {n-i \choose i}{l+i \choose 2i+1}=\sum_{i=0}^{l} {n-i \choose i-1}{l+i \choose 2i}.\] We will first address the…

组合数学 · 数学 2026-04-28 Flavien Mabilat

In this note we prove that \[ j!\,2^N \, \binom{N+j-1}{j} \, {}_2F_1\left(\begin{matrix}-j,-2j \\ -N-j+1 \end{matrix};-1\right) = \sum_{l=0}^N \binom{N}{l}\prod_{i=0}^{j-1}2(2i+1+l), \] where $ N $ and $ j $ are positive integers, which…

经典分析与常微分方程 · 数学 2024-01-19 Maxim L. Yattselev

Let k and n be positive integers. We mainly show that $$(ln+1) | k\binom{kn+ln}{kn},$$ $$2\binom{kn}n | \binom {2n}{n}C_{2n}^{(k-1)}$$, $$\binom{kn}n | (2k-1)C_n\binom{2kn}{2n},$$ $$\binom{2n}n | (k+1)C_n^{(k-1)}\binom{2kn}{kn},$$…

数论 · 数学 2010-06-01 Zhi-Wei Sun

In this paper we initiate the study of products and sums divisible by central binomial coefficients. We show that 2(2n+1)binom(2n,n)| binom(6n,3n)binom(3n,n) for every n=1,2,3,... Also, for any nonnegative integers $k$ and $n$ we have…

数论 · 数学 2010-05-06 Zhi-Wei Sun

Let $$ A_{m,n}(a)=\sum_{j=0}^m (-4)^j {m+j\choose 2j}\sum_{k=0}^{n-1} \sin(a+2k\pi/n) \cos^{2j}(a+2k\pi/n) $$ and $$ B_{m,n}(a)=\sum_{j=0}^m (-4)^j {m+j+1\choose 2j+1}\sum_{k=0}^{n-1} \sin(a+2k\pi/n) \cos^{2j+1}(a+2k\pi/n), $$ where $m\geq…

经典分析与常微分方程 · 数学 2023-01-02 Horst Alzer , Semyon Yakubovich

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…

数论 · 数学 2012-04-10 Victor J. W. Guo , Jiang Zeng

We prove that for any nonnegative integers $n$ and $r$ the binomial sum $$ \sum_{k=-n}^n\binom{2n}{n-k}k^{2r} $$ is divisible by $2^{2n-\min\{\alpha(n),\alpha(r)\}}$, where $\alpha(n)$ denotes the number of 1's in the binary expansion of…

组合数学 · 数学 2010-09-01 Hao Pan , Zhi-Wei Sun

Let $f_{n}=\sum_{i=0}^n \binom{n}{i}\binom{2n-2i}{n-i}$, $g_{n}= \sum_{i=1}^n \binom{n}{i}\binom{2n-2i}{n-i}$. Let $\{a_k\}_{k=1}$ be the set of all positive integers n, in increasing order, for which $\binom{2n}{n}$ is not divisible by 5,…

组合数学 · 数学 2013-02-04 Walter Shur

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…

经典分析与常微分方程 · 数学 2017-10-20 Horst Alzer , Semyon Yakubovich

Let $B_n$ be the $n$-th balancing number. In this paper, we give some explicit expressions of $\sum_{l=0}^{2 r-3}(-1)^l\binom{2 r-3}{l}\sum_{j_1+\cdots+j_r=n-2 l\atop j_1,\dots,j_r\ge 1}B_{j_1}\cdots B_{j_r}$ and…

数论 · 数学 2016-08-23 Takao Komatsu , Prasanta Kumar Ray

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…

数论 · 数学 2011-06-03 Zhi-Wei Sun , Roberto Tauraso

By using the Newton interpolation formula, we generalize the recent identities on the Catalan triangle obtained by Miana and Romero as well as those of Chen and Chu. We further study divisibility properties of sums of products of binomial…

数论 · 数学 2011-03-25 Victor J. W. Guo , Jiang Zeng

By polynomial (or extended binomial) coefficients, we mean the coefficients in the expansion of integral powers, positive and negative, of the polynomial $1+t +\cdots +t^{m}$; $m\geq 1$ being a fixed integer. We will establish several…

数论 · 数学 2016-07-26 Nour-Eddine Fahssi

By a very simple argument, we prove that if $l,m,n$ are nonnegative integers then $$\sum_{k=0}^l(-1)^{m-k}\binom{l}{k}\binom{m-k}{n}\binom{2k}{k-2l+m} =\sum_{k=0}^l\binom{l}{k}\binom{2k}{n}\binom{n-l}{m+n-3k-l}. On the basis of this…

组合数学 · 数学 2007-05-23 Hao Pan , Zhi-Wei Sun

In this paper we study the factors of some alternating sums of products of binomial and q-binomial coefficients. We prove that for all positive integers n_1,...,n_m, n_{m+1}=n_1, and 0\leq j\leq m-1, {n_1+n_{m}\brack…

数论 · 数学 2015-06-26 Victor J. W. Guo , Frederic Jouhet , Jiang Zeng

Using the expansion in a Fourier-Gegenbauer series, we prove several identities that extend and generalize known results. In particular, it is proved among other results, that \begin{equation*}…

经典分析与常微分方程 · 数学 2022-11-01 Omran Kouba

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…

数论 · 数学 2008-07-14 Zhi-Wei Sun
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