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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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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 ...
dzp's user avatar
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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 $...
მამუკა ჯიბლაძე's user avatar
2 votes
1 answer
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$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 $...
Notamathematician's user avatar
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 $...
Johann Cigler's user avatar
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 ...
Martin Rubey's user avatar
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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 ...
Samuel Johnston's user avatar
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+...
interstice's user avatar
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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 ...
Robert Wegner's user avatar
2 votes
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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 $C_{2n}=\frac1{2n+1}\binom{4n}{2n}$...
T. Amdeberhan's user avatar
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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} \...
T. Amdeberhan's user avatar
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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 ...
T. Amdeberhan's user avatar
10 votes
2 answers
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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?
banana's user avatar
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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 &...
Johann Cigler's user avatar
1 vote
0 answers
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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}}...
Johann Cigler's user avatar
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 ...
Per Alexandersson's user avatar
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,...
Johann Cigler's user avatar
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 ...
Aaron Li's user avatar
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,...
Johann Cigler's user avatar
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)^...
Roland Bacher's user avatar
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 ...
Joseph Iosue's user avatar
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 ...
GGT's user avatar
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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 $F_n(\mathbf{x})=\prod_{k=1}^n(x_1+\cdots+x_k)$ ...
T. Amdeberhan's user avatar
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-...
Per Alexandersson's user avatar
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 $s_1,\ldots,s_l$ ...
Roland Bacher's user avatar
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 ...
Svjetlan Feretic's user avatar
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 ...
T. Amdeberhan's user avatar
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 ...
GGT's user avatar
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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,...
eigengrau's user avatar
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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 ...
Johann Cigler's user avatar
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 $F=\sum_{n=0}^\infty C_n x^{n+1}$....
user196574's user avatar
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, $g(2n) = n$...
Notamathematician's user avatar
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 $d(k,n)= \det \left( C_{k + i + j} \right)_{i,j = 0}^{n - 1}=\prod_{i=1}^{k-1}\frac{\binom{2n+2j}{j}}{\binom{2j}{j}}$...
Johann Cigler's user avatar
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 ...
Mare's user avatar
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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 \...
Kacper Kurowski's user avatar
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 ...
Tuhin Mukherjee's user avatar
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 $...
Tuhin Mukherjee's user avatar
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 $x$-...
Sam Hopkins's user avatar
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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 ...
egunawan'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
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 ...
mlbaker's user avatar
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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 $\...
guacho's user avatar
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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 ...
T. Amdeberhan's user avatar
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]^...
T. Amdeberhan's user avatar
15 votes
1 answer
259 views

A formula for this generating function that is similar to the $qt$-Catalan numbers

I came up with the following conjecture: $$ \sum_{n \ge 0} z^n \sum_{\mu \vdash n} \frac{ t^{\sum l}q^{\sum a}}{\prod (q^a - t^{l+1})(t^l - q^{a+1})} = \exp\left(\sum_{n \ge 1} \frac{z^n}{n(q^n-1)(t^n-...
Drew's user avatar
  • 1,509
6 votes
0 answers
152 views

A conjecture about sums over partitions arising from Hilbert scheme of points

$\DeclareMathOperator{\leg}{\operatorname{leg}}\DeclareMathOperator{\arm}{\operatorname{arm}}$The following situation arose from the study of some localization computations on Hilbert schemes of ...
Drew's user avatar
  • 1,509
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 ...
Sam Hopkins's user avatar
  • 24.2k
13 votes
1 answer
497 views

What's the dimension of the Lie algebra generated by transpositions on $n$ objects?

Define a Lie bracket on the group algebra of the permutation group $S_n$ in the following way: $$[\sigma, \tau] = \sigma\circ\tau - \tau\circ\sigma,$$ where $\sigma, \tau \in S_n$, and the ...
WunderNatur's user avatar
4 votes
1 answer
469 views

A generalisation of the Catalan numbers

Let $n$ be a nonnegative integer. It is well-known that the number of lattice paths from $(0, 0)$ to $(n, n)$ with steps $(0, 1)$ and $(1, 0)$ that are never rising above the line $y=x$ is given by ...
Fikilis's user avatar
  • 155
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 ...
Per Alexandersson's user avatar
3 votes
0 answers
123 views

Seeking bijective proof of a recurrence for generalized Narayana numbers

Consider lattice paths in $d$ dimensions with the steps $X_1\mathrel{:=}(1,0,\dotsc,0)$, $X_2\mathrel{:=} (0,1,\dotsc,0)$,…, $X_d\mathrel{:=} (0,0,\dotsc,1)$. Let $\mathcal C(d, n)$ denote the set of ...
Tri's user avatar
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