# Catalan number

In combinatorial mathematics, the Catalan numbers form a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects. They are named after the Belgian mathematician Eugène Charles Catalan (1814–1894).

The n-th Catalan number is given directly in terms of binomial coefficients by

$C_{n}={\frac {1}{n+1}}{2n \choose n}={\frac {(2n)!}{(n+1)!\,n!}}=\prod \limits _{k=2}^{n}{\frac {n+k}{k}}\qquad {\text{for }}n\geq 0.$ The first Catalan numbers for n = 0, 1, 2, 3, ... are

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, ... (sequence A000108 in the OEIS).

## Properties

An alternative expression for Cn is

$C_{n}={2n \choose n}-{2n \choose n+1}={1 \over n+1}{2n \choose n}={1 \over n}{2n \choose n+1}\quad {\text{ for }}n\geq 0,$ because ${\tbinom {2n}{n+1}}={\tfrac {n}{n+1}}{\tbinom {2n}{n}}$ .

This shows that Cn is an integer, which is not immediately obvious from the first formula given. The last expression is valid due to the value of the reciprocal gamma function at 0, which cancels the 'division by zero'.

Another expression is

$C_{n}={\binom {2n-1}{n-1}}-{\binom {2n-1}{n-2}}$ because ${\tbinom {2n-1}{n-2}}={\tfrac {n-1}{n+1}}{\tbinom {2n-1}{n-1}}$ and ${\tbinom {2n-1}{n-1}}={\tfrac {1}{2}}{\tbinom {2n}{n}}$ .

The Catalan numbers satisfy the recurrence relations

$C_{0}=1\quad {\text{and}}\quad C_{n+1}=\sum _{i=0}^{n}C_{i}C_{n-i}\quad {\text{for }}n\geq 0$ and

$C_{0}=1\quad {\text{and}}\quad C_{n+1}={\frac {2(2n+1)}{n+2}}C_{n}$ which can be proved:

$C_{n+1}={\frac {1}{n+2}}{\binom {2n+2}{n+1}}={\frac {(2n+2)(2n+1)}{(n+2)(n+1)^{2}}}{\binom {2n}{n}}={\frac {2(2n+1)}{(n+2)(n+1)}}{\binom {2n}{n}}={\frac {2(2n+1)}{n+2}}C_{n}$ Asymptotically, the Catalan numbers grow as

$C_{n}\sim {\frac {4^{n}}{(n+1){\sqrt {n\pi }}}}\sim {\frac {4^{n}}{n^{3/2}{\sqrt {\pi }}}}\,,$ which can be shown by dividing the asymptote for the central binomial coefficients by $n+1$ , by using Stirling's approximation for $n!$ , or via generating functions.

The only Catalan numbers Cn that are odd are those for which n = 2k − 1; all others are even. The only prime Catalan numbers are C2 = 2 and C3 = 5.

The Catalan numbers have an integral representation

$C_{n}={\frac {1}{2\pi }}\int _{0}^{4}x^{n}{\sqrt {\frac {4-x}{x}}}\,dx\,$ .

This means that the Catalan numbers are a solution of a version of the Hausdorff moment problem.

The Catalan k-fold convolution is:

$\sum _{i_{1}+\cdots +i_{m}=n \atop i_{1},\ldots ,i_{m}\geq 0}C_{i_{1}}\cdots C_{i_{m}}={\begin{cases}{\dfrac {m(n+1)(n+2)\cdots (n+m/2-1)}{2(n+m/2+2)(n+m/2+3)\cdots (n+m)}}C_{n+m/2},&m{\text{ even}}\\[5pt]{\dfrac {m(n+1)(n+2)\cdots (n+(m-1)/2)}{(n+(m+3)/2)(n+(m+3)/2+1)\cdots (n+m)}}C_{n+(m-1)/2},&m{\text{ odd,}}\end{cases}}$ ## Applications in combinatorics

There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The book Enumerative Combinatorics: Volume 2 by combinatorialist Richard P. Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the cases C3 = 5 and C4 = 14.

• Cn is 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.
• Re-interpreting the symbol X as an open parenthesis and Y as a close parenthesis, Cn counts the number of expressions containing n pairs of parentheses which are correctly matched:
((()))     ()(())     ()()()     (())()     (()())
((ab)c)d     (a(bc))d     (ab)(cd)     a((bc)d)     a(b(cd))
• Successive applications of a binary operator can be represented in terms of a full binary tree, with each correctly matched bracketing describing an internal node. It follows that Cn is the number of full binary trees with n + 1 leaves:

Also, the interior of the correctly matching closing Y for the first X of a Dyck word contains the description of the left subtree, with the exterior describing the right subtree.

• Cn is the number of non-isomorphic ordered (or plane) trees with n + 1 vertices.
• Cn is the number of monotonic lattice paths along the edges of a grid with n × n square cells, which do not pass above the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up".

The following diagrams show the case n = 4:

This can be represented by listing the Catalan elements by column height: The dark triangle is the root node, the light triangles correspond to internal nodes of the binary trees, and the green bars are the leaves.
[0,0,0,0] [0,0,0,1] [0,0,0,2] [0,0,1,1]
[0,1,1,1] [0,0,1,2] [0,0,0,3] [0,1,1,2] [0,0,2,2] [0,0,1,3]
[0,0,2,3] [0,1,1,3] [0,1,2,2] [0,1,2,3]
• A convex polygon with n + 2 sides can be cut into triangles by connecting vertices with non-crossing line segments (a form of polygon triangulation). The number of triangles formed is n and the number of different ways that this can be achieved is Cn. The following hexagons illustrate the case n = 4:
• Cn is the number of stack-sortable permutations of {1, ..., n}. A permutation w is called stack-sortable if S(w) = (1, ..., n), where S(w) is defined recursively as follows: write wunv 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.
• Cn is the number of permutations of {1, ..., n} that avoid the permutation pattern 123 (or, alternatively, any of the other patterns of length 3); that is, the number of permutations with no three-term increasing subsequence. For n = 3, these permutations are 132, 213, 231, 312 and 321. For n = 4, they are 1432, 2143, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312 and 4321.
• Cn is the number of noncrossing partitions of the set {1, ..., n}. A fortiori, Cn never exceeds the nth Bell number. Cn is also the number of noncrossing partitions of the set {1, ..., 2n} in which every block is of size 2. The conjunction of these two facts may be used in a proof by mathematical induction that all of the free cumulants of degree more than 2 of the Wigner semicircle law are zero. This law is important in free probability theory and the theory of random matrices.
• Cn is the number of ways to tile a stairstep shape of height n with n rectangles. Cutting across the anti-diagonal and looking at only the edges gives full binary trees. The following figure illustrates the case n = 4:
• Cn is the number of ways to form a "mountain range" with n upstrokes and n downstrokes that all stay above a horizontal line. The mountain range interpretation is that the mountains will never go below the horizon.
• Cn is the number of standard Young tableaux whose diagram is a 2-by-n rectangle. In other words, it is the number of ways the numbers 1, 2, ..., 2n can be arranged in a 2-by-n rectangle so that each row and each column is increasing. As such, the formula can be derived as a special case of the hook-length formula.
• Cn is the number of ways that the vertices of a convex 2n-gon can be paired so that the line segments joining paired vertices do not intersect. This is precisely the condition that guarantees that the paired edges can be identified (sewn together) to form a closed surface of genus zero (a topological 2-sphere).
• Cn is the number of semiorders on n unlabeled items.
• In chemical engineering Cn−1 is the number of possible separation sequences which can separate a mixture of n components.
• Given an infinite complete binary decision tree and n-1 votes, $C_{n}$ is the number of possible voting outcomes, given that at any node you can split your votes anyway you want.

## Proof of the formula

There are several ways of explaining why the formula

$C_{n}={\frac {1}{n+1}}{2n \choose n}$ solves the combinatorial problems listed above. The first proof below uses a generating function. The other proofs are examples of bijective proofs; they involve literally counting a collection of some kind of object to arrive at the correct formula.

### First proof

We first observe that all of the combinatorial problems listed above satisfy Segner's recurrence relation

$C_{0}=1\quad {\text{and}}\quad C_{n+1}=\sum _{i=0}^{n}C_{i}\,C_{n-i}\quad {\text{for }}n\geq 0.$ For example, 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

$c(x)=\sum _{n=0}^{\infty }C_{n}x^{n}.$ The recurrence relation given above can then be summarized in generating function form by the relation

$c(x)=1+xc(x)^{2};$ in other words, this equation follows from the recurrence relation by expanding both sides into power series. On the one hand, the recurrence relation uniquely determines the Catalan numbers; on the other hand, the generating function relation can be algebraically solved to yield

$c(x)={\frac {1\pm {\sqrt {1-4x}}}{2x}}$ Choosing the minus sign, the fraction has a power series at 0 so its coefficients must therefore be the Catalan numbers. This solution satisfies

$\lim _{x\to 0^{+}}c(x)=C_{0}=1$ The square root term can be expanded as a power series using the identity

${\sqrt {1+y}}=\sum _{n=0}^{\infty }{{\frac {1}{2}} \choose n}y^{n}=\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}}{4^{n}(2n-1)}}{2n \choose n}y^{n}$ This is a special case of Newton's generalized binomial theorem; as with the general theorem, it can be proved by computing derivatives to produce its Taylor series.

Setting y = −4x gives

${\sqrt {1-4x}}=\sum _{n=0}^{\infty }{\frac {-1}{2n-1}}{2n \choose n}x^{n}$ and substituting this power series into the expression for c(x) and shifting the summation index n by 1, the expansion simplifies to

$\sum _{n=1}^{\infty }{\frac {1}{2(2n-1)}}{2n \choose n}x^{n-1}$ Let $N=n-1$ , so that

$\sum _{N=0}^{\infty }{\frac {1}{2(2N+1)}}{2N+2 \choose N+1}x^{N}$ and because ${\frac {1}{2(2N+1)}}{2N+2 \choose N+1}=C_{N}$ (see 'proof of recurrence' above)

we have

$c(x)=\sum _{n=0}^{\infty }{2n \choose n}{\frac {x^{n}}{n+1}}$ Another way to get c(x) is to solve for xc(x) and observe that $\int _{0}^{x}\!t^{n}\,dt$ appears in each term of the power series.

### Second proof Figure 1. The invalid portion of the path (dotted red) is flipped (solid red). Bad paths (after the flip) reach (n – 1, n + 1) instead of (n,n).

This proof uses André's reflection method, which was originally used in connection with Bertrand's ballot theorem. The reflection principle has been widely attributed to Désiré André, but his method did not actually use reflections and the reflection method is a variation due to Aebly and Mirimanoff.

We count the number of paths which start and end on the diagonal of a n × n grid. All such paths have n right and n up steps. Since we can choose which of the 2n steps are up or right, there are in total ${\tbinom {2n}{n}}$ monotonic paths of this type. A bad path crosses the main diagonal and touches the next higher (fatal) diagonal (red in the illustration).

The part of the path after the fatal touch is flipped about the fatal diagonal, as illustrated with the dotted line. This swaps all the right steps to up steps and vice versa. In the section of the path that is not reflected, there is one more up step than right steps, so therefore the remaining section of the bad path has one more right step than up steps. When this portion of the path is reflected, it will have one more up step than right steps.

Since there are still 2n steps, there are now n + 1 up steps and n − 1 right steps. So, instead of reaching (n,n), all bad paths after reflection end at (n − 1, n + 1). Because every monotonic path in the (n − 1) × (n + 1) grid meets the fatal diagonal, and because the reflection process is reversible, the reflection is therefore a bijection between bad paths in the original grid and monotonic paths in the new grid.

The number of bad paths is therefore:

${n-1+n+1 \choose n-1}={2n \choose n-1}={2n \choose n+1}$ and the number of Catalan paths (i.e. good paths) is obtained by removing the number of bad paths from the total number of monotonic paths of the original grid,

$C_{n}={2n \choose n}-{2n \choose n+1}={\frac {1}{n+1}}{2n \choose n}$ .

In terms of Dyck words, we start with a (non-Dyck) sequence of n X's and n Y's and interchange all X's and Y's after the first Y that violates the Dyck condition. After this Y, note that there is exactly one more Y than there are X's.

### Third proof

The following bijective proof, while being more involved than the previous one, provides a more natural explanation for the term n + 1 appearing in the denominator of the formula for Cn. A generalized version of this proof can be found in a paper of Rukavicka Josef (2011).

Suppose we are given a monotonic path, which may happen to cross the diagonal. The exceedance of the path is defined to be the number of vertical edges which lie above the diagonal. For example, in Figure 2, the edges lying above the diagonal are marked in red, so the exceedance of the path is 5.

If we are given a monotonic path whose exceedance is not zero, then we can apply the following algorithm to construct a new path whose exceedance is one less than the one we started with.

• Starting from the bottom left, follow the path until it first travels above the diagonal.
• Continue to follow the path until it touches the diagonal again. Denote by X the first such edge that is reached.
• Swap the portion of the path occurring before X with the portion occurring after X.

In Figure 3, the black dot indicates the point where the path first crosses the diagonal. The black edge is X, and we place the last lattice point of the red portion in the top-right corner, and the first lattice point of the green portion in the bottom-left corner, and place X accordingly, to make a new path, shown in the second diagram.

The exceedance has dropped from three to two. In fact, the algorithm will cause the exceedance to decrease by one, for any path that we feed it, because the first vertical step starting on the diagonal (at the point marked with a black dot) is the unique vertical edge that under the operation passes from above the diagonal to below it; all other vertical edges stay on the same side of the diagonal. Figure 4. All monotonic paths in a 3×3 grid, illustrating the exceedance-decreasing algorithm.

It is also not difficult to see that this process is reversible: given any path P whose exceedance is less than n, there is exactly one path which yields P when the algorithm is applied to it. Indeed, the (black) edge X, which originally was the first horizontal step ending on the diagonal, has become the last horizontal step starting on the diagonal. Alternatively, reverse the original algorithm to look for the first edge that passes below the diagonal.

This implies that the number of paths of exceedance n is equal to the number of paths of exceedance n − 1, which is equal to the number of paths of exceedance n − 2, and so on, down to zero. In other words, we have split up the set of all monotonic paths into n + 1 equally sized classes, corresponding to the possible exceedances between 0 and n. Since there are

${2n \choose n}$ monotonic paths, we obtain the desired formula

$C_{n}={\frac {1}{n+1}}{2n \choose n}.$ Figure 4 illustrates the situation for n = 3. Each of the 20 possible monotonic paths appears somewhere in the table. The first column shows all paths of exceedance three, which lie entirely above the diagonal. The columns to the right show the result of successive applications of the algorithm, with the exceedance decreasing one unit at a time. There are five rows, that is, C3 = 5.

### Fourth proof

This proof uses the triangulation definition of Catalan numbers to establish a relation between Cn and Cn+1.

Given a polygon P with n + 2 sides and a triangulation, mark one of its sides as the base, and also orient one of its 2n + 1 total edges. There are (4n + 2)Cn such marked triangulations for a given base.

Given a polygon Q with n + 3 sides and a (different) triangulation, again mark one of its sides as the base. Mark one of the sides other than the base side (and not an inner triangle edge). There are (n + 2)Cn + 1 such marked triangulations for a given base.

There is a simple bijection between these two marked triangulations: We can either collapse the triangle in Q whose side is marked (in two ways, and subtract the two that cannot collapse the base), or, in reverse, expand the oriented edge in P to a triangle and mark its new side.

Thus

$(4n+2)C_{n}=(n+2)C_{n+1}$ .

Write $\textstyle {\frac {4n-2}{n+1}}C_{n-1}=C_{n}$ .

Because $(2n)!=(2n)!!(2n-1)!!=2^{n}n!(2n-1)!!$ then

${\frac {(2n)!}{n!}}=2^{n}(2n-1)!!=(4n-2)!!!!$ Applying the recursion with $C_{1}=1$ gives the result.

### Fifth proof

This proof is based on the Dyck words interpretation of the Catalan numbers, so Cn is the number of ways to correctly match n pairs of brackets. We denote a (possibly empty) correct string with c and its inverse (where "[" and "]" are exchanged) with c'. Since any c can be uniquely decomposed into c = [ c1 ] c2, summing over the possible spots to place the closing bracket immediately gives the recursive definition

$C_{0}=1\quad {\text{and}}\quad C_{n+1}=\sum _{i=0}^{n}C_{i}\,C_{n-i}\quad {\text{for }}n\geq 0.$ Let b stand for a balanced string of length 2n—that is, containing an equal number of "[" and "]", so $\textstyle B_{n}={2n \choose n}$ . As before, any balanced string can be uniquely decomposed into either [ c ] b or ] c' [ b, so

$B_{n+1}=2\sum _{i=0}^{n}B_{i}C_{n-i}$ .

Any incorrect balanced string starts with c ], and the remaining string has one more [ than ], so

$B_{n+1}-C_{n+1}=\sum _{i=0}^{n}{2i+1 \choose i}C_{n-i}$ Also, from the definitions, we have:

$B_{n+1}-C_{n+1}=2\sum _{i=0}^{n}B_{i}C_{n-i}-\sum _{i=0}^{n}C_{i}\,C_{n-i}=\sum _{i=0}^{n}(2B_{i}-C_{i})C_{n-i}$ Therefore

$2B_{i}-C_{i}={2i+1 \choose i}={\frac {2i+1}{i+1}}B_{i}$ $C_{i}=B_{i}\left(2-{\frac {2i+1}{i+1}}\right)$ $C_{i}={\frac {1}{i+1}}{\binom {2i}{i}}$ ### Sixth proof

This proof is based on the Dyck words interpretation of the Catalan numbers and uses the cycle lemma of Dvoretzky and Motzkin.

We call a sequence of X's and Y's dominating if, reading from left to right, the number of X's is always strictly greater than the number of Y's. The cycle lemma states that any sequence of $m$ X's and $n$ Y's, where $m>n$ , has precisely $m-n$ dominating cyclic permutations. To see this, arrange the given sequence of $m+n$ X's and Y's in a circle and repeatedly remove adjacent pairs XY until only $m-n$ X's remain. Each of these X's was the start of a dominating cyclic permutation before anything was removed.

For example, consider $XXYXY$ . This is dominating, but none of its cyclic permutations $XYXYX$ , $YXYXX$ , $XYXXY$ and $YXXYX$ are.

In particular, when $m=n+1$ , there is exactly one dominating cyclic permutation. Removing the leading X from it (a dominating sequence must begin with X) leaves a Dyck sequence. Since there are ${2n+1 \choose n}$ in total, and each one belongs to one of 2n+1 equivalence classes, we have ${\frac {1}{2n+1}}{2n+1 \choose n}={\frac {1}{n+1}}{2n \choose n}=C_{n}$ distinct cycles of $n+1$ X's and $n$ Y's, each of which corresponds to exactly one Dyck sequence, hence $C_{n}$ counts Dyck sequences.

## Hankel matrix

The n×n Hankel matrix whose (ij) entry is the Catalan number Ci+j−2 has determinant 1, regardless of the value of n. For example, for n = 4 we have

$\det {\begin{bmatrix}1&1&2&5\\1&2&5&14\\2&5&14&42\\5&14&42&132\end{bmatrix}}=1.$ Moreover, if the indexing is "shifted" so that the (i, j) entry is filled with the Catalan number Ci+j−1 then the determinant is still 1, regardless of the value of n. For example, for n = 4 we have

$\det {\begin{bmatrix}1&2&5&14\\2&5&14&42\\5&14&42&132\\14&42&132&429\end{bmatrix}}=1.$ Taken together, these two conditions uniquely define the Catalan numbers.

Another feature unique to the Catalan-Hankel matrix is the determinant of the nxn submatrix starting at 2 has determinant n+1.

$\det {\begin{bmatrix}2\end{bmatrix}}=2$ $\det {\begin{bmatrix}2&5\\5&14\end{bmatrix}}=3$ $\det {\begin{bmatrix}2&5&14\\5&14&42\\14&42&132\end{bmatrix}}=4$ $\det {\begin{bmatrix}2&5&14&42\\5&14&42&132\\14&42&132&429\\42&132&429&1430\end{bmatrix}}=5$ et cetera.

## History Catalan numbers in Mingantu's book The Quick Method for Obtaining the Precise Ratio of Division of a Circle volume III

The Catalan sequence was 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 during his exploration of the Towers of Hanoi puzzle. The counting trick for Dyck words was found by Désiré André in 1887.

The name “Catalan numbers” originated from John Riordan.

In 1988, it came to light that the Catalan number sequence had been used in China by the Mongolian mathematician Mingantu by 1730. That is when he started to write his book Ge Yuan Mi Lu Jie Fa [The Quick Method for Obtaining the Precise Ratio of Division of a Circle], which was completed by his student Chen Jixin in 1774 but published sixty years later. Peter J. Larcombe (1999) sketched some of the features of the work of Mingantu, including the stimulus of Pierre Jartoux, who brought three infinite series to China early in the 1700s.

For instance, Ming used the Catalan sequence to express series expansions of sin(2α) and sin(4α) in terms of sin(α).

## Generalizations

The two-parameter sequence of non-negative integers ${\frac {(2m)!(2n)!}{(m+n)!m!n!}}$ is a generalization of the Catalan numbers. These are named super-Catalan numbers, by Ira Gessel. These number should not confused with the Schröder–Hipparchus numbers, which sometimes are also called super-Catalan numbers.

For $m=1$ , this is just two times the ordinary Catalan numbers, and for $m=n$ , the numbers have an easy combinatorial description. However, other combinatorial descriptions are only known for $m=2,3$ and $4$ , and it is an open problem to find a general combinatorial interpretation.

Sergey Fomin and Nathan Reading have given a generalized Catalan number associated to any finite crystallographic Coxeter group, namely the number of fully commutative elements of the group; in terms of the associated root system, it is the number of anti-chains (or order ideals) in the poset of positive roots. The classical Catalan number $C_{n}$ corresponds to the root system of type $A_{n}$ . The classical recurrence relation generalizes: the Catalan number of a Coxeter diagram is equal to the sum of the Catalan numbers of all its maximal proper sub-diagrams.