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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

5 votes
1 answer
129 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
231 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
  • 5,682
4 votes
0 answers
236 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 ...
Samuel Johnston's user avatar
8 votes
3 answers
844 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
0 votes
0 answers
83 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 ...
Robert Wegner's user avatar
2 votes
0 answers
166 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 $C_{2n}=\frac1{2n+1}\binom{4n}{2n}$...
T. Amdeberhan's user avatar
5 votes
0 answers
188 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} \...
T. Amdeberhan's user avatar
1 vote
0 answers
93 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 ...
T. Amdeberhan's user avatar
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?
banana's user avatar
  • 111
2 votes
0 answers
237 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 &...
Johann Cigler's user avatar
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}}...
Johann Cigler's user avatar
6 votes
0 answers
261 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
249 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
123 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
487 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

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