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# Catalan number

The Catalan numbers form a sequence of natural numbers that occur in various counting problems in combinatorics. The n-th Catalan number is defined by
$C_n = \frac{{2n\choose n}}{n+1} \qquad\mbox{ for }n\ge 0$
(see binomial coefficient). The first Catalan numbers for n=0,1,2,3,... are
1, 1, 2, 5, 14, 42, 132, 429, 1430,...
All Catalan numbers are natural numbers because one can verify
$C_n = {2n\choose n} - {2n\choose n-1} \qquad\mbox{ for }n\ge 1.$

Cn is equal to the number of Dyck words of length 2n. A Dyck word is a string consisting of n X's and n Y's such that no initial segment of the string has more Y's than X's. For example, the following are the Dyck words of length 6: XXXYYY, XYXXYY, XYXYXY, XXYYXY, XXYXYY. Accordingly, C3 = 5. Thinking of X as an open parenthesis and of Y as a closed one, every Dyck word of length 2n can be seen as an expression with n pairs of parentheses correctly placed together: ((())), ()(()), ()()(), (())(), (()()). Dyck words can be naturally represented as certain paths in a grid with n+1 by n+1 vertices connected by vertical and horizontal lines. The paths start in the lower left corner, end in the upper right corner, always move up or right, and never pass above the diagonal. X stands for "move right" and Y stands for "move up".

One can count Dyck words with the following trick due to D. André: focus on those words containing n X's and n Y's that are not Dyck words. In such a word, find the first Y that violates the Dyck condition, and then flip all letters following this Y from Y to X and vice versa. You obtain a word with n+1 Y's and n-1 X's, and in fact every word with n+1 Y's and n-1 X's can be obtained in this fashion in precisely one way. The number of these words is equal to

${2n\choose n-1}$
which is therefore the same as the number of non-Dyck words; the number of Dyck words must then be
${2n\choose n}-{2n\choose n-1}$
which is the n-th Catalan number Cn.

Cn is also the number of different ways n+1 factors can be completely parenthesized[?]. For n=3 for example, we have the following 5 different parenthesizations of 4 factors: a(b(cd)), a((bc)d), (ab)(cd), (a(bc))d, ((ab)c)d. Such expressions can be naturally represented as rooted ordered binary trees, so Cn also counts the number of these trees with n+1 leaves.

Cn is also equal to the number of different ways a polygon with n+2 sides can be cut into triangles by connecting vertices with straight lines.

If w is a finite sequence of different integers, we define a reordering S(w) of w recursively as follows: write w = unv where n is the largest element in w and u and v are shorter sequences, and set S(w) = S(u)S(v)n, with S being the identity for one-element sequences. A permutation w of {1,...,n} is called stack-sortable if S(w) = (1,...,n). The number of stack-sortable permutations of {1,...,n} is equal to Cn.

The Catalan numbers satisfy the recurrence relation

$C_0 = 1 \qquad \mbox{and} \qquad C_n=\sum_{i=0}^{n-1}C_i C_{n-1-i}\quad\mbox{for }n\ge 1$

This follows from the fact that every Dyck word w of length ≥ 2 can be written in a unique way in the form

w = Xw1Yw2
with (possibly empty) Dyck words w1 and w2.

The generating function for the Catalan numbers is defined by

$c(x)=\sum_{n=0}^\infty C_n x^n$
and using the above recurrence relation we see that
$c(x)=1+xc(x)^2\,$
and hence
$c(x) = \frac{1-\sqrt{1-4x}}{2x}$

Asymptotically, the Catalan numbers grow as

$C_n \sim \frac{4^n}{n^{3/2}\sqrt{\pi}}$
in the sense that the quotient of the n-th Catalan number and the expression on the right tends towards 1 for n → ∞.

The Catalan sequence was first described in the 18th century by Leonhard Euler, who was interested in the number of different ways of dividing a polygon into triangles. The sequence is named after Eugène Charles Catalan[?], who discovered the connection to parenthesized expressions. The counting trick for Dyck words was found by D. André in 1887.

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