View source for Catalan number

{{Use American English|date = March 2019}}
{{Short description|Recursive integer sequence}}
{{For|names of numbers in Catalan|List of numbers in various languages#Occitano-Romance}}
[[File:Noncrossing partitions 5.svg|thumb|The {{nowrap|1=C<sub>5</sub> = 42}} [[noncrossing partition]]s of a 5-element set (below, the other 10 of the [[Bell number|52]] [[Partition of a set|partitions]])]]
In [[combinatorics|combinatorial mathematics]], the '''Catalan numbers''' form a [[sequence]] of [[natural number]]s that occur in various [[Enumeration|counting problems]], often involving [[recursion|recursively]]-defined objects. They are named after the [[Belgium|Belgian]] [[mathematician]] [[Eugène Charles Catalan]] (1814–1894).

The ''n''th Catalan number is given directly in terms of [[binomial coefficient]]s by

:<math>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\ge 0.</math>

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

:[[1 (number)|1]], 1, [[2 (number)|2]], [[5 (number)|5]], [[14 (number)|14]], [[42 (number)|42]], [[132 (number)|132]], 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, ... {{OEIS|id=A000108}}.

== Properties ==
An alternative expression for ''C''<sub>''n''</sub> is
:<math>C_n = {2n\choose n} - {2n\choose n+1} = {1\over n+1}{2n\choose n} \quad\text{ for }n\ge 0,</math>
which is equivalent to the expression given above because <math>\tbinom{2n}{n+1}=\tfrac{n}{n+1}\tbinom{2n}n</math>. This shows that ''C''<sub>''n''</sub> is an [[integer]], which is not immediately obvious from the first formula given. This expression forms the basis for a [[#Second proof|proof of the correctness of the formula]].

The Catalan numbers satisfy the [[recurrence relation]]s<ref>{{cite article|first1=D.|last1=Bowman|first2=Alon|last2=Regev|journal=Adv. Appl. Math.|year=2014|volume=56|pages=35-55|doi=10.1016/j.aam.2014.01.004|title=Counting symmetry: classes of dissections of a convex regular polygon}}</ref>
:<math>C_0 = 1 \quad \text{and} \quad C_{n+1}=\sum_{i=0}^{n}C_i\,C_{n-i}\quad\text{for }n\ge 0,</math>
:<math> \sum_{i_1+\cdots+i_m=n, i_1,\ldots,i_m\ge 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}\\[5 pt]
   \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}</math>
and
:<math>C_0 = 1 \quad \text{and} \quad C_{n+1}=\frac{2(2n+1)}{n+2}C_n.</math>

Asymptotically, the Catalan numbers grow as

:<math>C_n \sim \frac{4^n}{n^{3/2}\sqrt{\pi}}</math>

in the sense that the quotient of the ''n''th Catalan number and the expression on the right [[Limit of a function|tends towards]] 1 as ''n'' approaches infinity.  This can be proved by using [[Stirling's approximation]] for&nbsp;''n''<nowiki>!</nowiki> or via generating functions; see the [[Generating function#Asymptotic growth of the Catalan numbers|Asymptotic growth of the Catalan numbers]] section of the [[Generating function]] article.

The only Catalan numbers ''C''<sub>''n''</sub> that are odd are those for which ''n''&nbsp;=&nbsp;2<sup>''k''</sup>&nbsp;−&nbsp;1; all others are even.  The only prime Catalan numbers are ''C''<sub>2</sub> = 2 and ''C''<sub>3</sub> = 5.<ref>{{Cite journal|title = Parity and primality of Catalan numbers |last = Koshy|first = Thomas |last2 = Salmassi|first2 = Mohammad |journal=The College Mathematics Journal|year=2006|volume=37|issue=1|pages = 52–53|url=https://www.maa.org/sites/default/files/Koshy-CMJ-2006.pdf}}</ref>

The Catalan numbers have an integral representation
:<math>C_n=\int_0^4x^n\rho(x)\,dx,</math>

where <math>\rho(x)=\tfrac 1{2\pi}\sqrt{\tfrac{4-x}{x}}.</math> This means that the Catalan numbers are a solution of the [[Hausdorff moment problem]] on the interval [0, 4] instead of [0, 1]. The [[orthogonal polynomials]] having the weight function <math>\rho(x)</math> on <math>[0, 4]</math> are
:<math>H_n(x)=\sum_{k=0}^n{n+k \choose n-k}(-x)^k.</math>

== 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 ''C''<sub>3</sub>&nbsp;=&nbsp;5 and ''C''<sub>4</sub>&nbsp;=&nbsp;14.

[[File:Dyck lattice D4.svg|thumb|Lattice of the 14 Dyck words of length 8 – ''('' and '')'' interpreted as ''up'' and ''down'']]
* ''C''<sub>''n''</sub> is the number of [[Dyck word]]s<ref>[http://www.FindStat.org/DyckPaths Equivalent definitions of Dyck paths]</ref> of length 2''n''. A Dyck word is a [[string (computer science)|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:
<div class="center"><big> XXXYYY &nbsp;&nbsp;&nbsp; XYXXYY &nbsp;&nbsp;&nbsp; XYXYXY &nbsp;&nbsp;&nbsp; XXYYXY &nbsp;&nbsp;&nbsp; XXYXYY.</big></div>
* Re-interpreting the symbol X as an open [[Bracket#Parentheses|parenthesis]] and Y as a close parenthesis, ''C''<sub>''n''</sub> counts the number of expressions containing ''n'' pairs of parentheses which are correctly matched:
<div class="center"><big> ((())) &nbsp;&nbsp;&nbsp; ()(()) &nbsp;&nbsp;&nbsp; ()()() &nbsp;&nbsp;&nbsp; (())() &nbsp;&nbsp;&nbsp; (()()) </big></div>
* ''C''<sub>''n''</sub> is the number of different ways ''n''&nbsp;+&nbsp;1 factors can be completely [[bracket|parenthesized]] (or the number of ways of [[associativity|associating]] ''n'' applications of a [[binary operator]]). For ''n'' = 3, for example, we have the following five different parenthesizations of four factors:
<div class="center"><big>((ab)c)d &nbsp;&nbsp;&nbsp; (a(bc))d &nbsp;&nbsp;&nbsp; (ab)(cd) &nbsp;&nbsp;&nbsp; a((bc)d) &nbsp;&nbsp;&nbsp; a(b(cd))</big></div>

[[File:Tamari lattice, trees.svg|thumb|The [[associahedron]] of order 4 with the C<sub>4</sub>=14 full binary trees with 5 leaves]]
* Successive applications of a binary operator can be represented in terms of a full [[binary tree]]. (A rooted binary tree is ''full'' if every vertex has either two children or no children.) It follows that ''C''<sub>''n''</sub> is the number of full binary [[tree (graph theory)|trees]] with ''n''&nbsp;+&nbsp;1 leaves:
[[Image:Catalan number binary tree example.png|center]]
* ''C''<sub>''n''</sub> is the number of non-isomorphic ordered trees with {{nowrap|''n'' + 1}} vertices. (An ordered tree is a rooted tree in which the children of each vertex are given a fixed left-to-right order.)<ref>Stanley p.221 example (e)</ref>
* ''C''<sub>''n''</sub> is the number of monotonic [[lattice path]]s 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:
[[Image:Catalan number 4x4 grid example.svg|450px|center]]
This can be succinctly represented by listing the Catalan elements by column height:<ref>{{cite journal|last=Črepinšek|first=Matej|last2=Mernik|first2=Luka|title=An efficient representation for solving Catalan number related problems|journal=International Journal of Pure and Applied Mathematics|year=2009|volume=56|issue=4|pages = 589–604|url=http://www.ijpam.eu/contents/2009-56-4/11/11.pdf}}</ref>

 <div style="text-align: center;">[0,0,0,0] [0,0,0,1] [0,0,0,2] [0,0,1,1]</div>

 <div style="text-align: center;">[0,1,1,1] [0,0,1,2] [0,0,0,3] [0,1,1,2] [0,0,2,2] [0,0,1,3]</div>

 <div style="text-align: center;">[0,0,2,3] [0,1,1,3] [0,1,2,2] [0,1,2,3]</div>

[[File:Tamari lattice, hexagons.svg|thumb|The triangles correspond to internal nodes of the binary trees.]]
* A [[convex polygon]] with ''n''&nbsp;+&nbsp;2 sides can be cut into [[triangle]]s by connecting vertices with non-crossing [[line segment]]s (a form of [[polygon triangulation]]). The number of triangles formed is ''n'' and the number of different ways that this can be achieved is ''C''<sub>''n''</sub>. The following hexagons illustrate the case ''n'' = 4:
[[Image:Catalan-Hexagons-example.svg|400px|center]]
* ''C''<sub>''n''</sub> is the number of [[Stack (data structure)|stack]]-sortable [[permutation]]s of {1, ..., ''n''}. A permutation ''w'' is called [[stack-sortable permutation|stack-sortable]] if ''S''(''w'') =&nbsp;(1,&nbsp;...,&nbsp;''n''), where ''S''(''w'') is defined recursively as follows: write ''w'' =&nbsp;''unv'' where ''n'' is the largest element in ''w'' and ''u'' and ''v'' are shorter sequences, and set ''S''(''w'') =&nbsp;''S''(''u'')''S''(''v'')''n'', with ''S'' being the identity for one-element sequences.
* ''C''<sub>''n''</sub> is the number of permutations of {1,&nbsp;...,&nbsp;''n''} that avoid the [[permutation pattern]]&nbsp;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''<sub>''n''</sub> is the number of [[noncrossing partition]]s of the set {1,&nbsp;...,&nbsp;''n''}. [[A fortiori argument|''A fortiori'']], ''C''<sub>''n''</sub> never exceeds the ''n''th [[Bell number]].  ''C''<sub>''n''</sub> is also the number of noncrossing partitions of the set {1,&nbsp;...,&nbsp;2''n''} 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'' [[cumulant]]s of degree more than&nbsp;2 of the [[Wigner semicircle law]] are zero. This law is important in [[free probability]] theory and the theory of [[random matrices]].
* ''C''<sub>''n''</sub> is the number of ways to tile a stairstep shape of height ''n'' with ''n'' rectangles. The following figure illustrates the case ''n''&nbsp;=&nbsp;4:
[[Image:Catalan stairsteps 4.svg|400px|center]]
* ''C''<sub>''n''</sub> 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.
[[File:Mountain Ranges.png|Catalan Number|400px|center]]
* ''C''<sub>''n''</sub> is the number of [[Young tableau#Tableaux|standard Young tableaux]] whose diagram is a 2-by-''n'' rectangle. In other words, it is the number of ways the numbers 1,&nbsp;2,&nbsp;...,&nbsp;2''n'' 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 [[Young tableau#Dimension of a representation|hook-length formula]].
* ''C''<sub>''n''</sub> is the number of ways that the vertices of a convex 2''n''-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).
* ''C''<sub>''n''</sub> is the number of [[semiorder]]s on ''n'' unlabeled items.<ref>{{citation
 | last1 = Kim | first1 = K. H.
 | last2 = Roush | first2 = F. W.
 | mr = 538212
 | issue = 2
 | journal = Journal of Combinatorics, Information &System Sciences
 | pages = 58–61
 | title = Enumeration of isomorphism classes of semiorders
 | volume = 3
 | year = 1978}}.</ref>
* In chemical engineering ''C''<sub>''n''−1</sub> is the number of possible separation sequences which can separate a mixture of ''n'' components.<ref>{{citation
 | last1 = Thompson | first1 = R. W.
 | last2 = King | first2 = C. J.
 | issue = 5
 | journal = American Institution of Chemical Engineers Journal
 | pages = 941–948
 | title = Systematic synthesis of separation schemes
 | volume = 18
 | year = 1972
 | doi=10.1002/aic.690180510}}.</ref>

== Proof of the formula ==

There are several ways of explaining why the formula
:<math>C_n = \frac{1}{n+1}{2n\choose n}</math>
solves the combinatorial problems listed above. The first proof below uses a [[generating function]]. The other proofs are examples of [[bijective proof]]s; 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 [[Johann Andreas Segner|Segner's]]<ref>A. de Segner, Enumeratio modorum, quibus figurae planae rectilineae per diagonales dividuntur in triangula. ''Novi commentarii academiae scientiarum Petropolitanae'' '''7''' (1758/59) 203–209.</ref> [[recurrence relation]]

:<math>C_0 = 1 \quad \text{and} \quad C_{n+1}=\sum_{i=0}^n C_i\,C_{n-i}\quad\text{for }n\ge 0.</math>

For example, every Dyck word ''w'' of length&nbsp;≥&nbsp;2 can be written in a unique way in the form
:''w'' = X''w''<sub>1</sub>Y''w''<sub>2</sub>
with (possibly empty) Dyck words ''w''<sub>1</sub> and ''w''<sub>2</sub>.

The [[generating function]] for the Catalan numbers is defined by

:<math>c(x)=\sum_{n=0}^\infty C_n x^n.</math>

The recurrence relation given above can then be summarized in generating function form by the relation

:<math>c(x)=1+xc(x)^2;</math>

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

:<math>c(x) = \frac{1-\sqrt{1-4x}}{2x}=\frac{2}{1+\sqrt{1-4x}}</math>

which has a power series at 0 and its coefficients must therefore be the Catalan numbers.  The chosen solution satisfies

:<math>\lim_{x \to 0^+} c(x) = C_0 = 1</math>

The other solution has a pole at 0 and this reasoning doesn't apply to it.

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

:<math>\sqrt{1+y} = \sum_{n=0}^\infty {\frac12 \choose n} y^n = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{4^n(2n-1)} {2n \choose n} y^n=1+\frac{1}{2}y-\frac{1}{8}y^2+\cdots.</math>

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'' = −4''x'' and substituting this power series into the expression for ''c''(''x'') and shifting the summation index ''n'' by 1, the expansion simplifies to

:<math>c(x) = \sum_{n=0}^\infty {2n \choose n} \frac{x^n}{n+1}.</math>

The coefficients are now the desired formula for ''C''<sub>''n''</sub>.

Another way to get ''c''(''x'') is to solve for ''xc''(''x'') and observe that <math>\int_0^x \! t^n \, dt</math> appears in each term of the power series.

=== Second proof ===

[[File:Catalan number-path reflection.svg|thumb|upright=0.5|Figure 1. The invalid portion of the path (in solid red color) is flipped. Bad paths reach (''n''&nbsp;–&nbsp;1,&nbsp;''n''&nbsp;+&nbsp;1) instead of (''n'',''n'').]]

This proof depends on a trick known as [[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.<ref>{{cite journal |url=http://webspace.ship.edu/msrenault/ballotproblem/monthly358-363-renault.pdf |last=Renault |first=Marc |title=Lost (and found) in translation: André's actual method and its application to the generalized ballot problem |journal=American Mathematical Monthly |volume=115 |year=2008 |issue=4 |pages=358–363 |doi=10.1080/00029890.2008.11920537 }}</ref>) We count the paths which start and end on the diagonal of the ''n''&nbsp;×&nbsp;''n'' grid. All such paths have  ''n'' rightward and ''n'' upward steps. Since we can choose which of the 2''n'' steps are upward (or, equivalently, rightward) ones, there are <math>\tbinom{2n}{n}</math> total monotonic paths of this type. A ''bad'' path will cross the main diagonal and touch the next higher (''fatal'') diagonal (depicted red in the illustration). We flip the portion of the path occurring after that touch about that fatal diagonal, as illustrated; this geometric operation amounts to interchanging all the rightward and upward steps after that touch. In the section of the path that is not reflected, there is one more upward step than rightward steps, so the remaining section of the bad path has one more rightward than upward step (because it ends on the main diagonal). When this portion of the path is reflected, it will also have one more upward step than rightward steps. Since there are still 2''n'' steps, there must now be ''n''&nbsp;+&nbsp;1 upward steps and ''n''&nbsp;−&nbsp;1 rightward steps. So, instead of reaching the target (''n'',''n''), all bad paths (after the portion of the path is reflected) will end in location (''n''&nbsp;−&nbsp;1,&nbsp;''n''&nbsp;+&nbsp;1). As any monotonic path in the (''n''&nbsp;−&nbsp;1)&nbsp;×&nbsp;(''n''&nbsp;+&nbsp;1) grid must meet the fatal diagonal, this reflection process sets up a bijection between the bad paths of the original grid and the monotonic paths of this new grid because the reflection process is reversible. The number of bad paths is therefore,

:<math>{n-1 + n+1 \choose n-1} = {2n \choose n-1} = {2n \choose n+1}</math>

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,

:<math>C_n = {2n \choose n} - {2n \choose n+1}.</math>

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. At that first Y, there are ''k''&nbsp;+&nbsp;1 Y's and ''k'' X's for some ''k'' between 1 and ''n''&nbsp;−&nbsp;1.

=== Third proof ===

The following bijective proof, while being more involved than the previous one, provides a more natural explanation for the term ''n''&nbsp;+&nbsp;1 appearing in the denominator of the formula for&nbsp;''C''<sub>''n''</sub>. A generalized version of this proof can be found in a paper of Rukavicka Josef (2011).<ref>Rukavicka Josef (2011), ''On Generalized Dyck Paths, Electronic Journal of Combinatorics [http://www.combinatorics.org/ojs/index.php/eljc/article/view/v18i1p40/pdf online]</ref>

[[Image:Catalan number exceedance example.png|frame|right|Figure 2. A path with exceedance 5.]]
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.

Now, if we are given a monotonic path whose exceedance is not zero, then we may 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''.
The following example should make this clearer. In Figure 3, the black dot indicates the point where the path first crosses the diagonal. The black edge is ''X'', and we swap the red portion with the green portion to make a new path, shown in the second diagram.
[[Image:Catalan number swapping example.png|frame|center|Figure 3. The green and red portions are being exchanged.]]
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.

[[Image:Catalan number algorithm table.png|frame|right|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.

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

:<math>{2n\choose n}</math>

monotonic paths, we obtain the desired formula

:<math>C_n = \frac{1}{n+1}{2n\choose n}.</math>

Figure 4 illustrates the situation for&nbsp;''n''&nbsp;=&nbsp;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,&nbsp;''C''<sub>3</sub>&nbsp;=&nbsp;5.

=== Fourth proof ===

This proof uses the triangulation definition of Catalan numbers to establish a relation between ''C<sub>n</sub>'' and ''C''<sub>''n''+1</sub>.  Given a polygon ''P'' with ''n''&nbsp;+&nbsp;2 sides, first mark one of its sides as the base.  If ''P'' is then triangulated, we can further choose and orient one of its 2''n''&nbsp;+&nbsp;1 edges.  There are (4''n''&nbsp;+&nbsp;2)''C''<sub>''n''</sub> such decorated triangulations.  Now given a polygon ''Q'' with ''n''&nbsp;+&nbsp;3 sides, again mark one of its sides as the base.  If ''Q'' is triangulated, we can further mark one of the sides other than the base side.  There are (''n''&nbsp;+&nbsp;2)''C''<sub>''n''&nbsp;+&nbsp;1</sub> such decorated triangulations.  Then there is a simple bijection between these two kinds of decorated triangulations:  We can either collapse the triangle in ''Q'' whose side is marked, or in reverse expand the oriented edge in ''P'' to a triangle and mark its new side.  Thus
:<math>(4n+2)C_n = (n+2)C_{n+1}.</math>
The binomial formula for ''C<sub>n</sub>'' follows immediately from this relation and the initial condition&nbsp;''C''<sub>1</sub>&nbsp;=&nbsp;1.

=== Fifth proof ===

This proof is based on the [[Dyck language|Dyck words]] interpretation of the Catalan numbers, so ''C''<sub>''n''</sub> 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''<sup>+</sup>. Since any ''c'' can be uniquely decomposed into ''c''&nbsp;=&nbsp;[&nbsp;''c''<sub>1</sub>&nbsp;] ''c''<sub>2</sub>, summing over the possible spots to place the closing bracket immediately gives the recursive definition
:<math>C_0 = 1 \quad \text{and} \quad C_{n+1} = \sum_{i=0}^n C_i\,C_{n-i}\quad\text{for }n\ge 0.</math>

Now let ''b'' stand for a ''balanced'' string of length 2''n''—that is, containing an equal number of "[" and "]"—and <math>\textstyle B_n = {2n\choose n} = d_n C_n</math> with some factor ''d''<sub>''n''</sub>&nbsp;≥&nbsp;1. As above, any balanced string can be uniquely decomposed into either [&nbsp;''c''&nbsp;]&nbsp;''b'' or ]&nbsp;''c''<sup>+</sup>&nbsp;[&nbsp;''b'', so
:<math>B_{n+1} = 2 \sum_{i=0}^n B_i C_{n-i}.</math>

Also, any incorrect balanced string starts with ''c''&nbsp;], so
:<math>B_{n+1} - C_{n+1} = \sum_{i=0}^n {2i+1 \choose i} C_{n-i} = \sum_{i=0}^n \frac{2i+1}{i+1} B_i C_{n-i}.</math>

Subtracting the above equations and using ''B''<sub>''i''</sub> = ''d''<sub>''i''</sub> ''C''<sub>''i''</sub> gives
:<math>C_{n+1} = 2 \sum_{i=0}^n d_i C_i C_{n-i} - \sum_{i=0}^n \frac{2i+1}{i+1} d_i C_i C_{n-i} = \sum_{i=0}^n \frac{d_i}{i+1} C_i C_{n-i}.</math>

Comparing coefficients with the original recursion formula for ''C''<sub>''n''</sub> gives ''d''<sub>''i''</sub> = ''i'' + 1, so
:<math>C_n = \frac{1}{n+1}{2n\choose n}.</math>

=== Sixth proof ===

This simple proof<ref>{{citation | last1 = Dershowitz | first1 = Nachum | last2 = Zaks | first2 = Shmuel | year =1980 | title = Enumerations of ordered trees| journal = Discrete Mathematics | volume = 31 | pages = 9–28. | doi=10.1016/0012-365x(80)90168-5}}</ref> is also based on the [[Dyck language|Dyck words]] interpretation of the Catalan numbers but uses the beautiful Cycle Lemma of Dvoretzky and Motzkin.<ref>{{citation | last1 =Dvoretzky| first1= Aryeh| last2= Motzkin| first2=Theodore |year=1947| title =A problem of arrangements|journal= Duke Mathematical Journal| volume = 14|pages = 305–313| doi=10.1215/s0012-7094-47-01423-3}}</ref>
Call a sequence of X's and Y's ''dominating'' if, reading from left to right, the ''imbalance'' is always positive, that is, the number of X's is always strictly greater than the number of Y's. The Cycle Lemma asserts that any sequence of <math>m</math> X's and <math>n</math> Y's, where <math>m> n</math>, has precisely <math>m-n</math> dominating cyclic permutations. To see this, just arrange the given sequence of <math>m+n</math> X's and Y's in a circle and repeatedly remove adjacent pairs XY until only <math>m-n</math> X's remain. Each of these X's was the start of a dominating cyclic permutation before anything was removed.
In particular, when  <math>m=n+1</math>, 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 <math>C_n=\frac{1}{2n+1}{2n+1 \choose n}=\frac{1}{n+1}{2n\choose n}</math> distinct cycles of <math>n+1</math> X's and <math>n</math> Y's, each of which corresponds to exactly one Dyck sequence, <math>C_n</math> counts Dyck sequences.

== Hankel matrix ==

The ''n''×''n'' [[Hankel matrix]] whose (''i'',&nbsp;''j'') entry is the Catalan number ''C''<sub>''i''+''j''−2</sub> has [[determinant]] 1, regardless of the value of ''n''. For example, for ''n'' = 4 we have
:<math>\det\begin{bmatrix}1 & 1 & 2 & 5 \\ 1 & 2 & 5 & 14 \\ 2 & 5 & 14 & 42 \\ 5 & 14 & 42 & 132\end{bmatrix} = 1.</math>

Moreover, if the indexing is "shifted" so that the (''i'', ''j'') entry is filled with the Catalan number ''C''<sub>''i''+''j''−1</sub> then the determinant is still 1, regardless of the value of ''n''.
For example, for ''n'' = 4 we have
:<math>\det\begin{bmatrix}1 & 2 & 5 & 14 \\ 2 & 5 & 14 & 42 \\ 5 & 14 & 42 & 132 \\ 14 & 42 & 132 & 429 \end{bmatrix} = 1.</math>

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

== History ==
[[File:Mingantu's Catalan numbers.JPG|thumb|400px|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 D. André in 1887.

In 1988, it came to light that the Catalan number sequence had been used in [[China]] by the Mongolian mathematician [[Mingantu]] by 1730.<ref>[https://www.math.ucla.edu/~pak/lectures/Cat/Larcombe-The_18th_century_Chinese_discovery_of_the_Catalan_numbers.pdf The 18th century Chinese discovery of the Catalan numbers]</ref><ref>[http://en.cnki.com.cn/Article_en/CJFDTOTAL-NMGX198802004.htm Ming Antu, the First Inventor of Catalan Numbers in the World]</ref> 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.  P.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 <math>\frac{(2m)!(2n)!}{(m+n)!m!n!}</math>
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 number]]s, which sometimes are also called super-Catalan numbers.

For <math>m=1</math>, this is just two times the ordinary Catalan numbers, and for <math>m=n</math>, the numbers have an easy combinatorial description.
However, other combinatorial descriptions are only known<ref name=Chen2012>{{cite arxiv|last1=Chen|first1=Xin|last2=Wang|first2=Jane|title=The super Catalan numbers S(m, m + s) for s ≤ 4|eprint=1208.4196}}</ref>
for <math>m=2</math> and <math>m=3</math>,
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 crystalographic [[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 <math>C_n</math> corresponds to the root system of type <math>A_n</math>. 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.<ref>[[Sergey Fomin]] and Nathan Reading, "Root systems and generalized associahedra", Geometric combinatorics, IAS/Park City Math. Ser. '''13''', [[American Mathematical Society]], Providence, RI, 2007, pp 63–131. {{arxiv|math/0505518}}</ref>

== See also ==
{{Portal|Mathematics}}
{{div col|colwidth=30em}}
* [[Associahedron]]
* [[Bertrand's ballot theorem]]
* [[Binomial transform]]
* [[Catalan's triangle]]
* [[Catalan–Mersenne number]]
* [[Fuss–Catalan number]]
* [[List of factorial and binomial topics]]
* [[Lobb numbers]]
* [[Narayana number]]
* [[Schröder–Hipparchus number]]
* [[Tamari lattice]]
* [[Wedderburn–Etherington number]]
{{div col end}}

== Notes ==
{{Reflist|30em}}

==References==
*  Stanley, Richard P. (2015), ''Catalan numbers''. Cambridge University Press, {{ISBN|978-1-107-42774-7}}.
* [[John H. Conway|Conway]] and [[Richard K. Guy|Guy]] (1996) ''The Book of Numbers''. New York: Copernicus, pp.&nbsp;96–106.
* {{citation
  | last = Gardner
  | first = Martin
  | authorlink = Martin Gardner
  | title = Time Travel and Other Mathematical Bewilderments
  | publisher = W.H. Freeman and Company
  | year = 1988
  | location = New York
  | pages = 253–266 (Ch. 20)
  | isbn = 0-7167-1924-X}}
* {{citation
  | last = Koshy
  | first = Thomas
  | title = Catalan Numbers with Applications
  | publisher = Oxford University Press
  | year = 2008
  | pages =
  | url = https://www.amazon.com/Thomas-Koshy/e/B001H6NZT4/ref=ntt_athr_dp_pel_1
  | isbn = 0-19-533454-X }}
* Koshy, Thomas & Zhenguang Gao (2011) "Some divisibility properties of Catalan numbers", [[Mathematical Gazette]] 95:96–102.
* {{cite journal | last1 = Larcombe | first1 = P.J. | year = 1999 | title =  The 18th century Chinese discovery of the Catalan numbers|url= https://www.math.ucla.edu/~pak/lectures/Cat/Larcombe-The_18th_century_Chinese_discovery_of_the_Catalan_numbers.pdf | journal = [[Mathematical Spectrum]] | volume = 32 | issue = | pages = 5–7 }}
* {{Citation | last1=Stanley | first1=Richard P. | title=Enumerative combinatorics. Vol. 2 | url=http://www-math.mit.edu/~rstan/ec/ | publisher=[[Cambridge University Press]] | series=Cambridge Studies in Advanced Mathematics | isbn=978-0-521-56069-6 | mr=1676282 | year=1999 | volume=62}}
* {{citation
  | last = Egecioglu
  | first = Omer
  | title = A Catalan–Hankel Determinant Evaluation
  | year = 2009
  | pages =
  | url = http://www.cs.ucsb.edu/~omer/DOWNLOADABLE/catalan_hankel09.pdf}}

==External links==
* {{citation|last=Stanley|first= Richard P.|url=http://www-math.mit.edu/~rstan/ec/catadd.pdf |format=PDF|title=Catalan addendum to Enumerative Combinatorics, Volume 2|year=1998}}
* {{mathworld|title=Catalan Number|urlname=CatalanNumber}}
* Dickau, Robert M.: [http://mathforum.org/advanced/robertd/catalan.html Catalan numbers] Further examples.
* Davis, Tom: [https://web.archive.org/web/20070216101521/http://mathcircle.berkeley.edu/BMC6/pdf0607/catalan.pdf Catalan numbers]. Still more examples.
* "Equivalence of Three Catalan Number Interpretations" from The Wolfram Demonstrations Project [http://demonstrations.wolfram.com/EquivalenceOfThreeCatalanNumberInterpretations/]
* {{wikiversity-inline|Partition related number triangles}}

{{Classes of natural numbers}}

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[[Category:Integer sequences]]
[[Category:Factorial and binomial topics]]
[[Category:Enumerative combinatorics]]
[[Category:Articles containing proofs]]