Catalan number

File:Noncrossing partitions 5.svg

The C₅ = 42 noncrossing partitions of a 5-element set (below, the other 10 of the 52 partitions)

The Catalan numbers are a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects. They are named after Eugène Catalan, though they were previously discovered in the 1730s by Minggatu.

The Template:Mvar-th Catalan number can be expressed directly in terms of the central binomial coefficients by

$C_n=\frac{1}{n+1}(\genfrac{}{}{0ex}{}{2n}{n})=\frac{(2n)!}{(n+1)!n!}\text{for }n\ge 0.$

The first Catalan numbers for n = 0, 1, 2, 3, ... are

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

Properties

An alternative expression for C_n is

$C_n=(\genfrac{}{}{0ex}{}{2n}{n})-(\genfrac{}{}{0ex}{}{2n}{n+1})$ for $n\ge 0,$

which is equivalent to the expression given above because $(\genfrac{}{}{0ex}{}{2n}{n+1})=\frac{n}{n+1}(\genfrac{}{}{0ex}{}{2n}{n})$. This expression shows that C_n is an integer, which is not immediately obvious from the first formula given. This expression forms the basis for a proof of the correctness of the formula.

Another alternative expression is

$C_n=\frac{1}{2n+1}(\genfrac{}{}{0ex}{}{2n+1}{n}),$

which can be directly interpreted in terms of the cycle lemma; see below.

The Catalan numbers satisfy the recurrence relations

$C_0=1\text{and}{C}_n=\sum _{i=1}^n{C}_{i-1}{C}_{n-i}\text{for }n>0$

and

$C_0=1\text{and}{C}_n=\frac{2(2n-1)}{n+1}{C}_{n-1}\text{for }n>0.$

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 Template:Mvar-th Catalan number and the expression on the right tends towards 1 as Template:Mvar approaches infinity.

This can be proved by using the asymptotic growth of the central binomial coefficients, by Stirling's approximation for $n!$, or via generating functions.

The only Catalan numbers C_n that are odd are those for which n = 2^k − 1; all others are even. The only prime Catalan numbers are C₂ = 2 and C₃ = 5. More generally, the multiplicity with which a prime Template:Mvar divides C_n can be determined by first expressing n + 1 in base Template:Mvar. For p = 2, the multiplicity is the number of 1 bits, minus 1. For Template:Mvar an odd prime, count all digits greater than (p + 1) / 2; also count digits equal to (p + 1) / 2 unless final; and count digits equal to (p − 1) / 2 if not final and the next digit is counted. The only known odd Catalan numbers that do not have last digit 5 are C₀ = 1, C₁ = 1, C₇ = 429, C₃₁, C₁₂₇ and C₂₅₅. The odd Catalan numbers, C_n for n = 2^k − 1, do not have last digit 5 if n + 1 has a base 5 representation containing 0, 1 and 2 only, except in the least significant place, which could also be a 3.

The Catalan numbers have the integral representations

$C_n=\frac{1}{2\pi }\underset{0}{\overset{4}{\int }}{x}^n\sqrt{\frac{4-x}{x}}dx=\frac{2}{\pi }{4}^n\underset{-1}{\overset{1}{\int }}{t}^{2n}\sqrt{1-t^2}dt.$

which immediately yields $\sum _{n=0}^{\mathrm{\infty }}\frac{C_n}{4^n}=2$.

This has a simple probabilistic interpretation. Consider a random walk on the integer line, starting at 0. Let −1 be a "trap" state, such that if the walker arrives at −1, it will remain there. The walker can arrive at the trap state at times 1, 3, 5, 7..., and the number of ways the walker can arrive at the trap state at time $2k+1$ is $C_k$. Since the 1D random walk is recurrent, the probability that the walker eventually arrives at −1 is $\sum _{n=0}^{\mathrm{\infty }}\frac{C_n}{2^{2n+1}}=1$.

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.

File:Dyck lattice D4.svg

Lattice of the 14 Dyck words of length 8 – Template:Mvar and Template:Mvar interpreted as up and down

• C_n is the number of Dyck words of length 2n. A Dyck word is a string consisting of Template:Mvar X's and Template:Mvar 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 up to length 6:

• Re-interpreting the symbol X as an opening parenthesis and Y as a closing parenthesis, C_n counts the number of expressions containing Template:Mvar pairs of parentheses which are correctly matched. For instance, for n = 3 these are

• C_n is the number of different ways n + 1 factors can be completely parenthesized, i.e. the number of ways of associating Template:Mvar applications of a binary operator (as in the matrix chain multiplication problem). For n = 3, for example, we have the following five different complete parenthesizations of four factors:

• Successive applications of a binary operator can be represented in terms of a full binary tree, by labeling each leaf a, b, c, d. It follows that C_n is the number of full binary trees with n + 1 leaves, or, equivalently, with a total of Template:Mvar internal nodes:

File:Catalan 4 leaves binary tree example.svg

File:Tamari lattice, trees.svg

The associahedron of order 4 with the C₄=14 full binary trees with 5 leaves

• C_n is the number of non-isomorphic ordered (or plane) trees with n + 1 vertices. See encoding ordered trees as binary trees. For example, C_n is the number of possible parse trees for a sentence (assuming binary branching), in natural language processing.
• C_n 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:

File:Catalan number 4x4 grid example.svg

This can be represented by listing the Catalan elements by column height:

File:Tamari lattice, hexagons.svg

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.

• 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 Template:Mvar and the number of different ways that this can be achieved is C_n. The following hexagons illustrate the case n = 4:

File:Catalan-Hexagons-example.svg

• C_n is the number of stack-sortable permutations of Template:Mset. A permutation Template:Mvar is called stack-sortable if S(w) = (1, ..., n), where S(w) is defined recursively as follows: write w = unv where Template:Mvar is the largest element in Template:Mvar and Template:Mvar and Template:Mvar are shorter sequences, and set S(w) = S(u)S(v)n, with Template:Mvar being the identity for one-element sequences.
• C_n is the number of permutations of Template:Mset 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.
• C_n is the number of noncrossing partitions of the set Template:Mset. A fortiori, C_n never exceeds the Template:Mvar-th Bell number. C_n is also the number of noncrossing partitions of the set Template:Mset in which every block is of size 2.
• C_n is the number of ways to tile a stairstep shape of height Template:Mvar with Template:Mvar rectangles. Cutting across the anti-diagonal and looking at only the edges gives full binary trees. The following figure illustrates the case n = 4:

File:Catalan stairsteps 4.svg

• C_n is the number of ways to form a "mountain range" with Template:Mvar upstrokes and Template:Mvar downstrokes that all stay above a horizontal line. The mountain range interpretation is that the mountains will never go below the horizon.

Mountain Ranges

$n=0:$	*	1 way
$n=1:$	/\	1 way
$n=2:$	0000000/\ /\/\,0/00\	2 ways
$n=3:$	0000000000000000000000000000000000/\ 00000000000/\0000/\000000/\/\0000/00\ /\/\/\,0/\/00\,0/00\/\,0/0000\,0/0000\	5 ways

• C_n is the number of standard Young tableaux whose diagram is a 2-by-Template:Mvar rectangle. In other words, it is the number of ways the numbers 1, 2, ..., 2n can be arranged in a 2-by-Template:Mvar 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.

123   124   125   134   135
456   356   346   256   246

• $C_n$ is the number of length Template:Mvar sequences that start with $1$, and can increase by either $0$ or $1$, or decrease by any number (to at least $1$). For $n=4$ these are $1234,1233,1232,1231,1223,1222,1221,1212,1211,1123,1122,1121,1112,1111$. From a Dyck path, start a counter at 0. An X increases the counter by 1 and a Y decreases it by 1. Record the values at only the X's. Compared to the similar representation of the Bell numbers, only $1213$ is missing.

Proof of the formula

There are several ways of explaining why the formula

$C_n=\frac{1}{n+1}(\genfrac{}{}{0ex}{}{2n}{n})$

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

$C_0=1\text{and}{C}_{n+1}=\sum _{i=0}^n{C}_i{C}_{n-i}\text{for }n\ge 0.$

For example, every Dyck word Template:Mvar 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

$c(x)=\sum _{n=0}^{\mathrm{\infty }}{C}_n{x}^n.$

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, interpreting xc² − c + 1 = 0 as a quadratic equation of Template:Mvar and using the quadratic formula, the generating function relation can be algebraically solved to yield two solution possibilities

$c(x)=\frac{1+\sqrt{1-4x}}{2x}$  or  $c(x)=\frac{1-\sqrt{1-4x}}{2x}$.

From the two possibilities, the second must be chosen because only the second gives

$C_0=\underset{x\to 0}{\lim }c(x)=1$.

The square root term can be expanded as a power series using the binomial series

$$\begin{array}{rl}1-\sqrt{1-4x}& =-\sum _{n=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{1/2}{n})(-4x{)}^n=-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}(2n-3)!!}{2^{n}n!}(-4x{)}^n\\ & =-\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(2n-1)!!}{2^{n+1}(n+1)!}(-4x{)}^{n+1}=\sum _{n=0}^{\mathrm{\infty }}\frac{2^{n+1}(2n-1)!!}{(n+1)!}{x}^{n+1}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{2(2n)!}{(n+1)!n!}{x}^{n+1}=\sum _{n=0}^{\mathrm{\infty }}\frac{2}{n+1}(\genfrac{}{}{0ex}{}{2n}{n})x^{n+1}.\end{array}$$ Thus, $$c(x)=\frac{1-\sqrt{1-4x}}{2x}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n+1}(\genfrac{}{}{0ex}{}{2n}{n})x^n.$$

Second proof

File:Catalan number-path reflection.svg

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).

Call a bad path one that starts at $(x,y)=(0,0)$, ends at $(n,n)$, is monotonic, and contains a point above the $y=x$ line. We count the number of bad paths by establishing a bijection with paths that start at $(0,0)$, end at $(n-1,n+1)$, and are monotonic.

For a given bad path, construct a reflected path as follows. Let $P$ be the first point on the bad path intersecting the line $y=x+1$. The bad path from $(0,0)$ to $P$ is the beginning of the reflected path. The part of the bad path from $P$ to $(n,n)$ reflected across the line $y=x+1$ is the rest of the reflected path. See the illustration for an example. The black line is the points shared between the two paths, the dotted red line is the rest of the bad path, the solid red line is the rest of the reflected path.

This is a bijection because every monotonic path from $(0,0)$ to $(n-1,n+1)$ is constructable from a bad path, and every reflected path is uniquely invertible by finding the unique point $P$, which must exist because every such path must intersect $y=x+1$.

The number of steps in the reflected path is $(n-1)+(n+1)=2n$. The number of upward steps is $n+1$ because the path is monotonic and starts at $y=0$ and ends at $y=n+1$.

The number of reflected paths can be counted in the usual way, by counting how many way upward steps may be distributed among total steps, which is

$(\genfrac{}{}{0ex}{}{2n}{n+1})$

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,

$C_n=(\genfrac{}{}{0ex}{}{2n}{n})-(\genfrac{}{}{0ex}{}{2n}{n+1})=\frac{1}{n+1}(\genfrac{}{}{0ex}{}{2n}{n}).$

This proof can be restated in terms of Dyck words. We start with a (non-Dyck) sequence of Template:Mvar X's and Template:Mvar Y's and interchange all X's and Y's after the first Y that violates the Dyck condition.

Third proof

This bijective proof provides a natural explanation for the term n + 1 appearing in the denominator of the formula for C_n. A generalized version of this proof can be found in a paper of Rukavicka Josef (2011).

File:Catalan number exceedance example.png

Figure 2. A path with exceedance 5.

Given a monotonic path, the exceedance of the path is defined to be the number of vertical edges above the diagonal. For example, in Figure 2, the edges above the diagonal are marked in red, so the exceedance of this path is 5.

Given a monotonic path whose exceedance is not zero, we apply the following algorithm to construct a new path whose exceedance is 1 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 Template:Mvar the first such edge that is reached.
• Swap the portion of the path occurring before Template:Mvar with the portion occurring after Template:Mvar.

In Figure 3, the black dot indicates the point where the path first crosses the diagonal. The black edge is Template:Mvar, 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.

File:Catalan number swapping example.png

Figure 3. The green and red portions are being exchanged.

The exceedance has dropped from 3 to 2. In fact, the algorithm causes the exceedance to decrease by 1 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 only vertical edge that changes from being above the diagonal to being below it when we apply the algorithm - all the other vertical edges stay on the same side of the diagonal.

File:Catalan number algorithm table.png

Figure 4. All monotonic paths in a 3×3 grid, illustrating the exceedance-decreasing algorithm.

It can be seen that this process is reversible: given any path Template:Mvar whose exceedance is less than Template:Mvar, there is exactly one path which yields Template:Mvar when the algorithm is applied to it. Indeed, the (black) edge Template:Mvar, 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 Template:Mvar 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 Template:Mvar. Since there are $(\genfrac{}{}{0ex}{}{2n}{n})$ monotonic paths, we obtain the desired formula $C_n=\frac{1}{n+1}(\genfrac{}{}{0ex}{}{2n}{n}).$

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, and the last column displays all paths no higher than the diagonal.

Using Dyck words, start with a sequence from $(\genfrac{}{}{0ex}{}{2n}{n})$. Let $X_d$ be the first Template:Mvar that brings an initial subsequence to equality, and configure the sequence as $(F)X_d(L)$. The new sequence is $LXF$.

Fourth proof

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

Given a polygon Template:Mvar 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)C_n such marked triangulations for a given base.

Given a polygon Template:Mvar 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)C_{n + 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 Template:Mvar whose side is marked (in two ways, and subtract the two that cannot collapse the base), or, in reverse, expand the oriented edge in Template:Mvar to a triangle and mark its new side.

Thus

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

Write $\frac{4n-2}{n+1}{C}_{n-1}=C_n.$

Because

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

we have

$\frac{(2n)!}{n!}=2^n(2n-1)!!=(4n-2)!!!!.$

Applying the recursion with $C_0=1$ gives the result.

Fifth proof

This proof is based on the Dyck words interpretation of the Catalan numbers, so $C_n$ is the number of ways to correctly match Template:Mvar pairs of brackets. We denote a (possibly empty) correct string with Template:Mvar and its inverse with Template:Mvar. Since any Template:Mvar can be uniquely decomposed into $c=(c_1)c_2$, summing over the possible lengths of $c_1$ immediately gives the recursive definition

$C_0=1\text{and}{C}_{n+1}=\sum _{i=0}^n{C}_i{C}_{n-i}\text{for }n\ge 0$.

Let Template:Mvar be a balanced string of length 2n, i.e. Template:Mvar contains an equal number of $($ and $)$, so $B_n=(\genfrac{}{}{0ex}{}{2n}{n})$. A balanced string can also be uniquely decomposed into either $(c)b$ or $)c^{\prime }(b$, so

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

Any incorrect (non-Catalan) balanced string starts with $c)$, and the remaining string has one more $($ than $)$, so

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

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, as this is true for all Template:Mvar,

$2B_i-C_i=(\genfrac{}{}{0ex}{}{2i+1}{i})$

$C_i=2B_i-(\genfrac{}{}{0ex}{}{2i+1}{i})$

$C_i=2(\genfrac{}{}{0ex}{}{2i}{i})-(\genfrac{}{}{0ex}{}{2i+1}{i})$

$C_i=\frac{1}{i+1}(\genfrac{}{}{0ex}{}{2i}{i})$

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 circular shifts. To see this, arrange the given sequence of $m+n$ X's and Y's in a circle. Repeatedly removing XY pairs leaves exactly $m-n$ X's. Each of these X's was the start of a dominating circular shift before anything was removed. For example, consider ${X}{X}{Y}{X}{Y}$. This sequence is dominating, but none of its circular shifts ${X}{Y}{X}{Y}{X}$, ${Y}{X}{Y}{X}{X}$, ${X}{Y}{X}{X}{Y}$ and ${Y}{X}{X}{Y}{X}$ are.

A string is a Dyck word of $n$ X's and $n$ Y's if and only if prepending an X to the Dyck word gives a dominating sequence with $n+1$ X's and $n$ Y's, so we can count the former by instead counting the latter. In particular, when $m=n+1$, there is exactly one dominating circular shift. There are $(\genfrac{}{}{0ex}{}{2n+1}{n})$ sequences with exactly $n+1$ X's and $n$ Y's. For each of these, only one of the $2n+1$ circular shifts is dominating. Therefore there are $\frac{1}{2n+1}(\genfrac{}{}{0ex}{}{2n+1}{n})=C_n$ distinct sequences of $n+1$ X's and $n$ Y's that are dominating, each of which corresponds to exactly one Dyck word.

Hankel matrix

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

$\det \left[\begin{array}{cccc}1& 1& 2& 5\\ 1& 2& 5& 14\\ 2& 5& 14& 42\\ 5& 14& 42& 132\end{array}\right]=1.$

Moreover, if the indexing is "shifted" so that the (i, j) entry is filled with the Catalan number C_i+j−1 then the determinant is still 1, regardless of the value of Template:Mvar. For example, for n = 4 we have

$\det \left[\begin{array}{cccc}1& 2& 5& 14\\ 2& 5& 14& 42\\ 5& 14& 42& 132\\ 14& 42& 132& 429\end{array}\right]=1.$

Taken together, these two conditions uniquely define the Catalan numbers.

Another feature unique to the Catalan–Hankel matrix is that the n × n submatrix starting at 2 has determinant n + 1.

$\det \left[\begin{array}{c}2\end{array}\right]=2$

$\det \left[\begin{array}{cc}2& 5\\ 5& 14\end{array}\right]=3$

$\det \left[\begin{array}{ccc}2& 5& 14\\ 5& 14& 42\\ 14& 42& 132\end{array}\right]=4$

$\det \left[\begin{array}{cccc}2& 5& 14& 42\\ 5& 14& 42& 132\\ 14& 42& 132& 429\\ 42& 132& 429& 1430\end{array}\right]=5$

et cetera.

History

File:Mingantu's Catalan numbers.JPG

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 1751 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 reflection counting trick (second proof) 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 $\sin (2\alpha )$ and $\sin (4\alpha )$ in terms of $\sin (\alpha )$.

Generalizations

The Catalan numbers can be interpreted as a special case of the Bertrand's ballot theorem. Specifically, $C_n$ is the number of ways for a candidate A with n + 1 votes to lead candidate B with Template:Mvar votes.

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, per Ira Gessel. These 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.

The Catalan numbers are a solution of a version of the Hausdorff moment problem.

For coprime positive integers Template:Mvar and Template:Mvar, the rational Catalan numbers $\frac{1}{r+s}(\genfrac{}{}{0ex}{}{r+s}{r})$ count the number of lattice paths with steps of unit length rightwards and upwards from (0,0) to (r,s) that never go above the line ry = sx.

Catalan k-fold convolution

The Catalan Template:Mvar-fold convolution is:

$\sum _{\genfrac{}{}{0ex}{}{i_1+\cdots +i_k=n}{i_1,\dots ,i_k\ge 0}}{C}_{i_1}\cdots C_{i_k}=\frac{k}{2n+k}(\genfrac{}{}{0ex}{}{2n+k}{n})$

See also

• Associahedron
• Bertrand's ballot theorem
• Binomial transform
• Catalan's triangle
• Catalan–Mersenne number
• Delannoy number
• Fuss–Catalan number
• List of factorial and binomial topics
• Lobb numbers
• Motzkin number
• Narayana number
• Narayana polynomials
• Schröder number
• Schröder–Hipparchus number
• Semiorder
• Tamari lattice
• Wedderburn–Etherington number
• Wigner's semicircle law

References

• Stanley, Richard P. (2015), Catalan numbers. Cambridge University Press, Template:ISBN.
• Conway and Guy (1996) The Book of Numbers. New York: Copernicus, pp. 96–106.
• Koshy, Thomas & Zhenguang Gao (2011) "Some divisibility properties of Catalan numbers", Mathematical Gazette 95:96–102.

External links

• Davis, Tom: Catalan numbers. Still more examples.
• "Equivalence of Three Catalan Number Interpretations" from The Wolfram Demonstrations Project [1]
• Template:Wikiversity-inline