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Questions tagged [catalan-numbers]

The Catalan numbers form the sequence of numbers starting 1,1,2,5,14,42,... with explicit formula $\frac{1}{n+1}\binom{2n}{n}$. It counts many combinatorial objects like planar binary trees, triangulations, noncrossing partitions, Dyck paths, etc. See https://oeis.org/A000108

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2 votes
0 answers
89 views

The linear independence and linear elimination of non-crossing matching polynomials

Consider the polynomial set: $$ f_{ij} = (t_i - t_j)x_i x_j + x_i - x_j, \quad (1 \leq j < i \leq 2n) $$ where $ t_1, t_2, \dots, t_{2n} $ are pairwise distinct. Let's look at the non-crossing ...
6 votes
0 answers
164 views

Can one naturally transform Tamari lattices into distributive lattices with the same number of elements?

Many of the zillions of combinatorial objects counted by Catalan numbers come with various lattice structures. The $n$th Tamari lattice $T_n$, as originally defined, is the lattice of all those maps $...
6 votes
0 answers
214 views

Looking for a combinatorial proof for an identity involving $q$-Catalan triangles

Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Following my earlier post on MO, one fine colleague asked me if there is a $q$-analogue of the identity formed by the so-called Shapiro's ...
2 votes
1 answer
147 views

$R$-recursion for Fibonacci numbers using signed Catalan numbers

Let $F_n$ be A000045 (i.e., Fibonacci numbers). Here $$ F_n = F_{n-1} + F_{n-2}, \\ F_0 = 0, F_1 = 1. $$ Let $C_n$ be A000108 (i.e., Catalan numbers). Here $$ C_n = \frac{1}{n+1}\binom{2n}{n}. $$ Let $...
5 votes
1 answer
148 views

Identities for the generating functions of a sort of convolution powers of the Narayana numbers

Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers. It satisfies $$\frac{1}{c(x)^k}+x^k c(x)^k=L_k(1,-x),$$ where $L_n(x,s)$ denote the Lucas polynomials defined by $...
2 votes
0 answers
235 views

Injection of Catalan objects into 3-connected planar graphs

Let $C_n = \frac{1}{n+1}\binom{2n}{n}$ be the $n$-th Catalan number, counting, for example, the number of (rooted) triangulations of the $(n+2)$-gon. Let $P_n$ be the number of three-connected planar ...
3 votes
1 answer
331 views

Even more generalized Catalan numbers

What is the number of ways to parenthesize $n$ elements using applications of operators of arbitrary arities larger than or equal to $2$? For example, for $n=3$, there are $3$ ways: $$ abc, a(bc),(ab)...
4 votes
0 answers
272 views

Representations of $\mathrm{sl}(3,\mathbb{C})$ and Catalan-like paths

Background on representations of $\mathrm{sl}(3,\mathbb{C})$ In Chapter 6 of Brian C. Hall's book "Lie Groups, Lie Algebras, and Representations", he constructs the irreducible ...
8 votes
3 answers
921 views

Alternating Sum Involving Catalan Numbers

I was wondering if anyone knew how to obtain a simpler closed form of the following sum(or had any other insights regarding it): $$\sum_{k=0}^n (-1)^k{n \choose k} C_{2n-2-k} $$ Here $C_n = \frac{1}{n+...
0 votes
0 answers
92 views

Solving a Catalan-like recursion of polynomials, related to the KdV energies

I am working on a PDE problem. The goal is to connect the higher order energies of the Gross-Pitaevskii equation to those of the Korteweg-de-Vries equation. As these higher order energies are ...
2 votes
0 answers
172 views

Lattice paths avoiding holes

Consider lattice paths from $(0,0)$ to $(2n,2n)$ with steps $N=(0,1)$ and $E=(1,0)$ avoiding the points $(2i-1,2i-1)$ for all $1\leq i\leq n$. There are Catalan many $C_{2n}=\frac1{2n+1}\binom{4n}{2n}$...
5 votes
0 answers
190 views

Yet, another generalization of Catalan determinants

The discussion on this page is motivated by Johann Cigler's MO question. My intention arose from a possible generalization of Cigler's matrix $$A_{n,m}=\left( \binom{2m}{j-i+m}-\binom{2m}{m-i-j-1} \...
10 votes
2 answers
2k views

Proving an identity about Catalan numbers

$$C_{n} = \sum_{i=1}^n (-1)^{i-1} \binom{n-i+1}{i} C_{n-i}$$ Are there any good combinatorial proofs or algebraic proofs of this?
1 vote
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100 views

Super Catalan (super ballot) numbers

We refer to this article by Ira Gessel. In section 6, page 10, equation (28), the Super Catalan numbers are defined as $$S(m,n)=\frac{(2m)!(2n)!}{m!n!(m+n)!}.$$ On page 12, equation (31), there goes ...
2 votes
0 answers
241 views

Determinants of band matrices which are related to Hankel matrices of Catalan numbers

Let $A_{n,m}$ be the band matrix $$ A_{n,m}=\left( \binom{2m}{j-i+m}-\binom{2m}{m-i-j-1} \right)_{0\leq {i,j} \leq {n-1}}.$$ For example, $$A_{6,2}=\left ( \begin{matrix} 2 & 3 & 1& 0 &...
1 vote
0 answers
78 views

Shifted Hankel determinants for convolutions of Catalan numbers

It is well known that for $m\in \mathbb N$ the Hankel determinants $$D_m(n)= \det\left(C_{i+j+m}\right)_{0\leq i,j\leq {n-1}}$$ satisfy $D_m(n)=p_m(n)$, where $p_m(n)=\prod_{1 \leq i \leq j \leq {m-1}}...
4 votes
2 answers
346 views

Equidistribution of returns and height of first peak of Dyck paths

I believe that it is "well known" that the following two statistics on Dyck paths have symmetric joint distribution: number of returns to the axis $RET(D)$ height of the first peak (or length of the ...
7 votes
0 answers
252 views

Hankel determinants for some convolutions of Catalan numbers

Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers and let $$x^k c(x)^{2k}=(c(x)-1)^k =\sum_{n\geq0}c(k,n)x^n.$$ Consider the determinants $$D(k,n,m)= \det\left(c(k,...
6 votes
0 answers
295 views

Catalan numbers from matchings?

There are several examples of interpreting the Catalan numbers as non-nesting or non-crossing matchings of some graph. My question is: Is there a family of graphs $G_1,G_2,\dotsc$ with the number of ...
5 votes
1 answer
502 views

A polynomial identity related to Catalan numbers

Let $F_n^{(k)}(x)= \sum_j {\binom{n+(k-1)j}{kj} x^j}$ and $G_n^{(k)}(x)= \sum_j {\binom{n+j}{kj} x^j}.$ I am interested in the coefficients ${a_{n,k,j}}$ such that $$G_n^{(k)}(x)=\sum_{j\geq0 }{a_{n,...
3 votes
1 answer
131 views

Why is this alternating sum involving Catalan numbers $\sum_{i=0}^{\lfloor t/2 \rfloor} (-1)^{i+1} \binom{t-i}{i} C_{t-i-1} = 0$ for all $t$?

I need the result that for all $t$, $$\sum_{i=0}^{\lfloor t/2 \rfloor} (-1)^{i+1} \binom{t-i}{i} C_{t-i-1} = 0,$$ where $C_{t-i-1}$ is the $(t-i-1)$-th Catalan number. I've checked for $t$ up to ...
7 votes
1 answer
608 views

Reciprocity for fans of bounded Dyck paths

This is a continuation of some questions asked by Johann Cigler: Number of bounded Dyck paths with "negative length" and Number of bounded Dyck paths with negative length as Hankel ...
7 votes
2 answers
424 views

A sequence of polynomials related to Catalan numbers

The sequence of polynomials $$P_n=\sum_{k=0}^{\lfloor(2n-1)/3\rfloor} \frac{(2n-2k-1)!(2n-2k-2)!}{k!(n-k)!(n-k-1)!(2n-3k-1)!}x^k$$ satisfies apparently the identities $$0=\sum_{j=0}^nP_{n-j}(P_j-(-x)^...
16 votes
1 answer
545 views

Conjecture on sum over permutations of products of Catalan numbers

Context In a recent paper involving entanglement in linear optics, we came across some summations involving Catalan numbers and permutations. In particular, these sums arise when doing integration ...
6 votes
0 answers
112 views

Bijection between forests and skew SYT + Cyclic sieving

Consider the two-row skew shape $\lambda_n = (2n+1,n)/(1)$. The number of standard Young tableaux of this shape is $\binom{3n}{n}-\binom{3n}{n-2}$ (since one can easily biject this to the set of non-...
4 votes
0 answers
205 views

Non-crossing and crossing bijection in higher genus

This is a follow-up question of my SO post I'll briefly mention it here. So given a $n$ cycle say $(1,2,\ldots,n)$, what are the monotonic 2 -tuples, of the form $(a,b)(c,d)$, monotonicity in on the ...
2 votes
1 answer
141 views

Counting monomials and $q$-Catalan polynomials

Define $N(F)$ to be the number of monomials of a multi-variable polynomial $F$. For example $N(x^2y+3xy-y^5)=3$. If $\mathbf{x}=(x_1,\dots,x_n)$ and $F_n(\mathbf{x})=\prod_{k=1}^n(x_1+\cdots+x_k)$ ...
4 votes
1 answer
221 views

Reference for a definition of Catalan numbers

The $l$-th Catalan number ${2l\choose l}\frac{1}{l+1}$ is equal to the number of sequences $s_0,\ldots,s_{l+1}$ of length $l+2$ with the following properties: (1) $s_0=s_{l+1}=1$ and $s_1,\ldots,s_l$ ...
3 votes
1 answer
193 views

Sequences that sums up to second differences of Bell and Catalan numbers

Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$. Let $g(n)$ be A025480, $g(2n) = n$...
1 vote
1 answer
139 views

A big list of Narayana-enumerated objects

By a Narayana-enumerated object I mean an object whose count is given by the Narayana number $N(n,k)=\frac{1}{n} {n \choose k} {n \choose k-1}$. Can you give me a reference to some good big list of ...
5 votes
1 answer
347 views

Counting monomials and the Catalan numbers

Given a multi-variable polynomial $F$, denote the number of monomials by $N(F)$. Take for instance, \begin{align*}N(x(x+y)+(x+y)^2-(x-y)^2)=N(x^2+5xy)&=2 \qquad \text{and} \\ N((x+z)(x+y)^2)=N(x^3 ...
2 votes
1 answer
199 views

Sequence of monotone tuples and permutation condition for rotation

I was doing some counting in $S_n$ symmetric group I encountered the following problem, which also someway related to central factorial number. So given a $n$ cycle say $(1,2,\ldots,n)$, what are the ...
0 votes
1 answer
148 views

What is this numerically-generated function?

This question is an "outgrowth" of https://math.stackexchange.com/questions/4380919/ which led to a numerically-generated two-parameter function $f_b(n)$, where $b$ is the number base $2,3,4,...
33 votes
3 answers
2k views

A double grading of catalan numbers

This is something I found in trying to work on Vince Vatter's excellent question. I have no solution, but a much more precise conjecture. Recall that a rooted planar tree is a rooted tree where, for ...
5 votes
1 answer
251 views

Hankel determinants for q-Catalan numbers where q is a root of unity?

Let ${C_n}(q)$ be the weight of the Dyck paths of semilength $n$ where the upsteps have weight $1$ and the downsteps which end on height $i$ have weight $q^i$. They satisfy ${C_n}(q) = \sum\limits_{j ...
6 votes
1 answer
281 views

Another generalization of parity of Catalan numbers

Recently, a question by T. Amdeberhan gathered up many enjoyable proofs that a Catalan number $C_n$ is odd if and only if $n=2^r-1$. Noam D. Elkies' answer considered $F=\sum_{n=0}^\infty C_n x^{n+1}$....
4 votes
1 answer
423 views

Generating functions for Hankel determinants of Catalan numbers

The Hankel determinants of the Catalan numbers are well known and can be written as $d(k,n)= \det \left( C_{k + i + j} \right)_{i,j = 0}^{n - 1}=\prod_{i=1}^{k-1}\frac{\binom{2n+2j}{j}}{\binom{2j}{j}}$...
11 votes
2 answers
367 views

Kernel of a matrix and the Catalan numbers

Let $B_n$ denote the Boolean lattice of a set with $n \geq 2$ elements and $C_n$ the matrix with entries $c_{i,j}=1$ if $i \leq j$ and $c_{i,j}=0$ else, where $i,j\in B_n$. Let $M_n:=C_n+C_n^T$ (this ...
4 votes
1 answer
298 views

Generalization of Catalan numbers

Some time ago I was trying to find a closed form formula for the number of tuples $(a_k)_{k=1}^{n+s}$ of non-negative integers satisfying following conditions: $\sum_{k=1}^{n+s} a_k = n$, $\forall m \...
17 votes
1 answer
885 views

Proof of certain $q$-identity for $q$-Catalan numbers

Let us use the standard notation for $q$-integers, $q$-binomials, and the $q$-analog $$ \operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q. $$ I want to prove that for all ...
2 votes
0 answers
182 views

Does this question have anything to do with Catalan numbers?

I think this question has something to do with Catalan numbers but I'm not really sure. I want to find out the number of strings that consist of $n$ $L$'s and $n$ $R$'s, under the constraint that for ...
0 votes
1 answer
287 views

Dyck words and Catalan numbers

One of the many applications of the $n$th Catalan number is to calculate the number of strings consisting of $n$ $X$'s and $n$ $Y$'s, such that any prefix of the string will contain at least as many $...
6 votes
0 answers
381 views

Reference request: colored Motzkin path interpretation of Catalan numbers

Recall that a Dyck path of length $2n$ is a lattice path in $\mathbb{Z}^2$ from $(0,0)$ to $(2n,0)$ consisting of $n$ up steps $U=(1,1)$ and $n$ down steps $D=(1,-1)$ which never goes below the $x$-...
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^...
5 votes
1 answer
416 views

Reference request: recurrence relation for Catalan numbers

I would like to know if the following recurrence relation for Catalan numbers (see mathoverflow.net/questions/191524 and also math.stackexchange.com/questions/2113830) has appeared in a paper or a ...
3 votes
1 answer
366 views

A recurrence relation on Catalan numbers

In the classical problem of bracketing $n$ numbers, I know the response is $C_{n-1}$. I find this $$C_{n-1}=\sum_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}(-1)^{i+1}\binom{n-i}{i}C_{n-1-i}$$ but I ...
6 votes
1 answer
362 views

Are the “generalized Catalan numbers” of Dumitrescu–Mulase the "moments" of some "multivariate Wigner semicircle distribution"?

The classical Catalan numbers $$ C_n = \frac{1}{n+1} \binom{2n}{n}, $$ well-known for their numerous combinatorial interpretations (the second volume of Stanley's Enumerative Combinatorics famously ...
7 votes
0 answers
184 views

Upper bounds for a sequence of integers

Given $\alpha\geq0$ we consider the sequence $$ C_k=k^\alpha\sum_{j=0}^{k-1}C_jC_{k-1-j} $$ with $C_0=1$. I'm interested in upper bounds (in terms of $\alpha$) for such a sequence. I know that when $\...
5 votes
1 answer
2k views

What does the $q$-Catalan Numbers count?

I had completed a paper describing the $q$-Catalan numbers, which is the $q$-analog of the Catalan numbers. The $n$-th Catalan numbers can be represented by: $$C_n=\frac{1}{n+1}{2n \choose n}$$ and ...
13 votes
5 answers
2k views

Looking for a combinatorial proof for a Catalan identity

Let $C_n=\frac1{n+1}\binom{2n}n$ be the familiar Catalan numbers. QUESTION. Is there a combinatorial or conceptual justification for this identity? $$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^...