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6 votes
2 answers
719 views

Recreation with Catalan

Consider the well-known sequence $C_k=\frac1{k+1}\binom{2k}k$ of Catalan numbers. I came across the below identity while working with certain generating functions. I thought it might be of interest to ...
T. Amdeberhan's user avatar
3 votes
3 answers
756 views

Ordinary partitions vs partitions into odd parts

Let $\mathcal{P}(n)$ be the set of all unrestricted partitions of $n$ while $\mathcal{O}(n)$ stand for the set of all partitions of $n$ into odd parts. We adopt the power notation for partitions $\...
T. Amdeberhan's user avatar
3 votes
1 answer
215 views

Seeking for a combinatorial argument for partition identities

Given an integer partition $\lambda$, introduce the following quantities: \begin{align*} c(\lambda)&=\sum_{i\geq1}\left\lceil\frac{\lambda_i}2\right\rceil, \qquad c_o(\lambda)=\sum_{i\geq1}\left\...
T. Amdeberhan's user avatar
7 votes
1 answer
287 views

A reference for a sum found in Gould's Combinatorial Identities book

On p. 49 in Gould's book Combinatorial Identities, the author states that the sum $$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{2n}{2k}^{-1}$$ "... arises naturally in a statistical problem; it ...
Sela Fried's user avatar
15 votes
2 answers
1k views

A rather curious identity on sums over triple binomial terms

While exploring the Baxter sequences from my earlier MO post, I obtained a rather curious identity (not listed on OEIS either). I usually try to employ the Wilf-Zeilberger (WZ) algorithm to justify ...
T. Amdeberhan's user avatar
4 votes
0 answers
208 views

Extract this constant term

Given a Laurent polynomial $F$ in the variables $\mathbf{t}=(t_1,\dots,t_n)$, let $CT_{\vec{\mathbf{t}}}\,F$ denote its constant term. For example, $CT_{t_1,t_2}((8t_1-\frac1{3t_1t_2})(5t_1t_2+t_2^2+\...
T. Amdeberhan's user avatar
4 votes
0 answers
97 views

"Convolving" a general Catalan with classical Catalan

Consider what is sometimes known as generalized Catalan sequence $$\mathcal{{\color{red}C}}_{a,b}:=\frac{2b+1}{a+b+1}\binom{2a}{a+b}.$$ Observe that $\mathcal{{\color{red}C}}_{n,0}$ reduces to the ...
T. Amdeberhan's user avatar
10 votes
2 answers
484 views

Identity involving a quadratic term inside the Pochhammer symbol

This identity came up in my research: $$ \sum_{m=1}^n m^2 \frac{(\frac{xy}n + m-1)_{2m-1} (n+m-1)_{2m-1}}{(x+m)_{2m+1} (y+m)_{2m+1}} = \frac{n^2}{(x^2-n^2) (y^2 - n^2)}. $$ Here $n$ is a fixed ...
Anton Mellit's user avatar
  • 3,772
1 vote
1 answer
220 views

Gaussian at $q=\pm1$, log-concave polynomials, Catalan numbers

Let $[n]_q!=\prod_{j=1}^n\frac{1-q^j}{1-q}$ with $[0]_q!:=1$ and the Gaussian polynomials $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\,\cdot\,[n-k]_q!}$. Adopt the convention that $\binom{n}k_q=0$ whenever $k&...
T. Amdeberhan's user avatar
2 votes
1 answer
220 views

$q$-binomial sum, slightly

Recall that $[n]_{q}!=\prod_{j=1}^n\frac{1-q^{j}}{1-q}$ and $\binom{n}k_{q}=\frac{[n]_{q}!}{[k]_{q}![n-k]_{q}!}$. Then the $q$-binomial theorem states $$\sum_{k=0}^n\binom{n}k_qq^{\binom{k}2}=\prod_{k=...
T. Amdeberhan's user avatar
6 votes
2 answers
217 views

A convolution-type identity for the "major index"

For a permutation $\pi\in\frak{S}_n$, define the number of descents of $\pi$ as $$\text{des}(\pi)=\vert\{i: \pi(i)>\pi(i+1)\}\vert.$$ The following is a well-known (and interesting) identity: $$\...
T. Amdeberhan's user avatar
6 votes
5 answers
944 views

Combinatorial proof of Catalan's identity

Consider the problem of tiling a board of length $n$ with squares of size $1×1$ and dominoes of size $1×2$, Let's denote $f_n$ to be the number of ways to tile this so-called ($n$)-board.Then $f_n=F_{...
user avatar
6 votes
2 answers
1k views

Products and sum of cubes in Fibonacci

Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$. Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a ...
T. Amdeberhan's user avatar
7 votes
1 answer
325 views

Looking for a $q$-analogue of a binomial identity

The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae): $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \...
T. Amdeberhan's user avatar
3 votes
1 answer
186 views

Is there a $q$-analogue to Shapiro's convolution identity?

Let $C_n=\frac1{n+1}\binom{2n}n$ denote the Catalan numbers. This question is motivated by the (unanswered) MO post by Alexander Burstein and my own (answered by Fedor Petrov) MO post. Specifically, ...
T. Amdeberhan's user avatar
10 votes
0 answers
349 views

A bijective proof for the odd companion to Shapiro's Catalan convolution

Shapiro's Catalan convolution is the following formula (where $C_n$ is the $n$th Catalan number): $$ \sum_{k=0}^{n}{C_{2k}C_{2(n-k)}}=4^nC_n. $$ In other words, letting $C(z)=\sum_{n=0}^{\infty}{C_nz^...
Alexander Burstein's user avatar
1 vote
2 answers
211 views

Bilinear recurrence relation between even Bernoulli numbers

Throughout this question $n$ is a positive integer greater than 1. Consider the following well-known identity by Euler, $$\sum_{k=1}^{n-1} \binom{2n}{2k}B_{2k}B_{2n-2k}=-(2n+1)B_{2n}.$$ Rather ...
bryanjaeho's user avatar
8 votes
1 answer
914 views

A special binomial identity in need of a proof

I've encountered a curious identity as a codicil in some work. Is there a proof or reference? $$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\frac{x^k}{1+x^{2k+1}}=\frac{x^n}{1+x^{2n+1}}.$$
T. Amdeberhan's user avatar
8 votes
2 answers
325 views

A link between hooks and contents: Part II

This is a question in the spirit of an earlier problem. Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. Recall also the notation for the content of a cell $...
T. Amdeberhan's user avatar
11 votes
2 answers
1k views

Proofs of some combinatorial identities

Just wondering if anyone knows any references in the literature to bijections corresponding to the following simple generating function identities. Let $B(z)=\dfrac{1}{\sqrt{1-4z}}$ and $C(z)=\dfrac{1-...
Alexander Burstein's user avatar
2 votes
2 answers
273 views

Alternating binomial-harmonic sum: evaluation request

Let $H_k=\sum_{j=1}^k\frac1j$ be the harmonic numbers. QUESTION. Can you find an evaluation of the following sum? $$\sum_{a=1}^b(-1)^a\binom{n}{b-a}\frac{H_{b-a}}a.$$
T. Amdeberhan's user avatar
1 vote
1 answer
154 views

$q$-plane partitions & specialization & interlinks

MacMahon's enumeration of all plane partions (PP) inside an $n$-cube generalizes to $${\tt PP_n}(q)=\prod_{i,j,k=1}^n\frac{1-q^{i+j+k-1}}{1-q^{i+j+k-2}}.$$ A $q$-analogue of symmetric plane partitions ...
T. Amdeberhan's user avatar
3 votes
2 answers
696 views

Open problems and known identities involving sums

As many people here, I know of a few identities involving expressions of the type $\sum_{i}\ f(i)$, with "arbitrarily complicated $f(\cdot)$", as well as closed formulas in some cases. I also know ...
Anthony Labarre's user avatar