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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
6
votes
Accepted
Why are the dinv-statistic and the partition length equidistributed?
This formula appears in Exercise 1.103 of Enumerative Combinatorics, vol. 1, second ed. It was first proved by K. Liu, C. H. F. Yan, and J. Zhou, Sci. China, Ser. A 45 (2002), 420-431. A combinatorial …
12
votes
Accepted
Curious Catalan convolutions
For the first identity, see additional problem A33 in my book Catalan Numbers. References are given to bijective proofs by Andrews and Nagy.
6
votes
Kernel of a matrix and the Catalan numbers
Another suggestion: let $A_{2n}=M_{2n}-2I$ ($I$=identity matrix), so we are
interested in $\ker A_{2n}$. Let $V$ be the vector space on which $A_{2n}$
acts, so we can regard $V$ as having a basis cons …
12
votes
Catalan numbers as sums of squares of numbers in the rows of the Catalan triangle - is there...
Another way of looking at it is that $m_{n,n-2i}$ is the number of
standard Young tableaux of shape $(n-i,i)$. An elegant proof that
$\sum_k m_{n,k}^2=C_n$ goes back to MacMahon. Take a standard Young …
7
votes
Coincidences between average Catalan tableaux
This is not a solution, but rather a long comment. Let
$f^{a,b}$ denote the number of standard Young tableaux (SYT)
of shape $(a,b)$. The number of SYT $T$ of shape $(n,n)$
with $T_{1d}=k$ is $f^{d-1, …