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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
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
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
17 votes
1 answer
886 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
30 votes
1 answer
1k views

Mysterious symmetry - in search for a bijection

I have a mysterious symmetry that I have not managed to prove. First some definitions (see picture below) Fix a partition that fit in a staircase shape with $n$ rows. There are $Catalan(n)$ such ...
Per Alexandersson's user avatar
11 votes
2 answers
604 views

Does $q$-Catalan number count subspaces?

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\...
Pritam Majumder's user avatar
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 ...
Gong D's user avatar
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15 votes
5 answers
2k views

enumerative meaning of natural q-Catalan numbers

Define $[n]=(1-q^n)/(1-q)$ and $[n]!=[1][2][3] \cdots [n]$, so that $[2n]!/[n]![n+1]!$ is a polynomial in $q$ (the most algebraically natural $q$-analogue of the Catalan numbers); what enumerative ...
James Propp's user avatar
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