Related papers: New developments of an old identity

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.

We know [Rui Duarte and Ant\'onio Guedes de Oliveira, New developments of an old identity, manuscript arXiv:1203.5424, submitted.] that, for every non-negative integer numbers $n,i,j$ and for every real number $\ell$, $$ \sum_{i+j=n}…

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…

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*}…

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…

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…

In this paper we use a formula for the $n$-th power of a $2\times2$ matrix $A$ (in terms of the entries in $A$) to derive various combinatorial identities. Three examples of our results follow. 1) We show that if $m$ and $n$ are positive…

For any $m,n\in\mathbb{N}$ we first give new proofs for the following well known combinatorial identities \begin{equation*} S_n(m)=\sum\limits_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k^m}=\sum\limits_{n\geq r_1\geq r_2\geq...\geq r_m\geq…

Let $[x]$ be the greatest integer not exceeding $x$. In the paper we introduce the sequence $\{U_n\}$ given by $U_0=1$ and $U_n=-2\sum_{k=1}^{[n/2]}\binom n{2k}U_{n-2k}\quad(n\ge 1)$, and establish many recursive formulas and congruences…

We survey combinatorial interpretations of some dozen identities for the double factorial such as, for instance, (2n-2)!! + Sum_{k=2}^{n} (2n-1)!!(2k-4)!!/(2k-1)!! = (2n-1)!!. Our methods are mostly bijective.

Based on an interesting identity of Bat{\i}r we derive new identities for double sums involving famous number sequences. We also prove some double sum identities for binomial transform pairs.

We obtain a three-parameter $q$-series identity that generalizes two results of Chan and Mao. By specializing our identity, we derive new results of combinatorial significance in connection with $N(r, s, m, n)$, a function counting certain…

We provide bijective proofs of two classic identities that are very simple to prove using generating functions, but surprisingly difficult to prove combinatorially. The problem of finding a bijective proof for the first identity was first…

In this paper we study the sum $$\sum_{j_1+j_2+...+j_d=n}\prod_{i=1}^d F_{k\cdot j_i},$$ where $d\geq2$ and $k\geq1$.

This note provide bijective proofs of two combinatorial identities involving generalized Catalan number $C_{m,5}(n)={m\over 5n+m}{5n+m\choose n}$ recently proposed by Sun.

We prove an interesting identity for the sum of determinants, which is a generalization of the sum of a geometric progression. The proof is quite long and a number of other identities are proved along the way. Some of the more elementary…

We find the joint distribution of three simple statistics on lattice paths of n upsteps and n downsteps leading to a triple sum identity for the central binomial coefficient {2n}-choose-{n}. We explain why one of the constituent double sums…

The title identity appeared as Problem 75-4, proposed by P. Barrucand, in Siam Review in 1975. The published solution equated constant terms in a suitable polynomial identity. Here we give a combinatorial interpretation in terms of card…

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…

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…