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Tagged with catalan-numbers binomial-coefficients
9 questions
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 ...
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 ...
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 ...
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?
$$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^...
3
votes
1
answer
235
views
Reference request for some determinants of binomial coefficients
Let $C_{n}=\binom{2n}{n}\frac{1}{n+1}$ be a Catalan number. I am interested in books or papers where the following identities occur:
$$\det\left(\binom{i+j+1}{i-j+1}\right)_{0 \leq i,j\leq {n-1}}=C_{n}...
12
votes
1
answer
553
views
A matrix identity related to Catalan numbers
Let $$C_n=\frac{1}{2n+1}\binom{2n+1}{n}$$ be a Catalan number. It is well-known that $$(\sum_{n\ge{0}}C_n x^n)^k=\sum_{n\ge{0}}C(n,k)x^n$$ with $$C(n,k)=\frac{k}{2n+k}\binom{2n+k}{n}.$$
It is also ...
9
votes
0
answers
213
views
Some quotients of Hankel determinants
This question has been inspired by Hankel determinants of binomial coefficients.
For a sequence $\{h_{n}\}_{n=0}^{\infty}$ denote by $H_n$ the Hankel matrix
$$H_{n}:=\begin{pmatrix}
h_{0} & h_{...
4
votes
2
answers
9k
views
Combinatorial proof of a recurrence for the Catalan numbers
I would like to ask whether there is a combinatorial proof of the following recurrence relation for Catalan numbers:
$$
C_{n+1}=\frac{4n+2}{n+2} C_n.
$$
Thanks!~