Catalan number

From Academic Kids

The Catalan numbers, named after the Belgian mathematician Eugne Charles Catalan (1814–1894), form a sequence of natural numbers that occur in various counting problems in combinatorics, many of which have a recursive flavour. The nth Catalan number is given in terms of binomial coefficients by the simple formula

<math>C_n = \frac{1}{n+1}{2n\choose n} \qquad\mbox{ for }n\ge 0.<math>

Properties of the Catalan numbers

One can verify that an alternative expression for Cn is

<math>C_n = {2n\choose n} - {2n\choose n-1} \qquad\mbox{ for }n\ge 1.<math>

This shows that Cn is a natural number, which is not a priori obvious from the first formula given. This expression forms the basis for André's proof of the correctness of the formula (see below under second proof).

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

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452...

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 tends towards 1 for n → ∞. (This can be proved by using Stirling's approximation for n!.)

Applications in combinatorics

There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The book Enumerative Combinatorics: Volume 2 (Cambridge University Press, 1999) by combinatorialist Richard Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the case C3 = 5.

  • 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:
  • 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:
((()))     ()(())     ()()()     (())()     (()())
  • Cn is the number of different ways n + 1 factors can be completely parenthesized. For n = 3 for example, we have the following five different parenthesizations of four factors:
a(b(cd))     a((bc)d)     (ab)(cd)     (a(bc))d     ((ab)c)d
  • Cn is the number of monotonic paths along the edges of a grid with n × n square cells, which do not cross 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 = 3:
Missing image

  • Cn is the number of rooted ordered binary trees with n + 1 leaves:
Missing image

  • Cn is the number of non-isomorphic full binary trees with n vertices that have children, usually called internal vertices or branches. (A rooted binary tree is full if every vertex has either two children or no children.)
  • Cn is the number of different ways a convex polygon with n + 2 sides can be cut into triangles by connecting vertices with straight lines. The following pentagons illustrate the case n = 3:
Missing image

  • 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 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.

Proof of the formula

There are several ways of explaining why the formula given for Cn is correct; that is, why it solves the combinatorial problems listed above. The first proof below uses a generating function, and is not particularly illuminating. The second and third 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: using generating functions

The Catalan numbers satisfy the recurrence relation

<math>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.<math>

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

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

and using the above recurrence relation we see that


and hence

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

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

<math>\sqrt{1+y} = 1 - 2\sum_{n=1}^\infty {2n-2 \choose n-1} \left(\frac{-1}{4}\right)^n \frac{y^n}{n} ,<math>

which can be proved, for example, by the binomial theorem, together with judicious juggling of factorials. Substituting this into the above expression for c(x) produces, after further manipulation,

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

Equating coefficients yields the desired formula for Cn.

Second proof

This proof depends on a trick due to D. André, which is now more generally known as the reflection principle (not to be confused with the Schwarz reflection theorem in complex analysis). It is most easily expressed in terms of the "monotonic paths which do not cross the diagonal" problem (see above).

Missing image
Figure 1. The green portion of the path is being flipped.

Suppose we are given a monotonic path in an n × n grid that does cross the diagonal. Find the first edge in the path that lies above the diagonal, and flip the portion of the path occurring after that edge, along a line parallel to the diagonal. (In terms of Dyck words, we are starting with a sequence of n X's and n Y's which is not a Dyck word, and exchanging all X's with Y's after the first Y that violates the Dyck condition.) The resulting path is a monotonic path in an (n − 1) × (n + 1) grid. Figure 1 illustrates this procedure; the green portion of the path is the portion being flipped.

Since every monotonic path in the (n − 1) × (n + 1) grid must cross the diagonal at some point, every such path can be obtained in this fashion in precisely one way. The number of these paths is equal to

<math>{2n\choose n-1}<math>.

Therefore, to calculate the number of monotonic n × n paths which do not cross the diagonal, we need to subtract this from the total number of monotonic n × n paths, so we finally obtain

<math>{2n\choose n}-{2n\choose n-1}<math>

which is the nth Catalan number Cn.

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.

Missing image
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 pairs of 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 circle 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.

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

Notice that 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.

Missing image
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.

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

<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 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. Since there are five rows, C3 = 5.


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 Eugne Charles Catalan, who discovered the connection to parenthesized expressions. The counting trick for Dyck words was found by D. André in 1887.


  • Stanley, R.P. (1999): Enumerative Combinatorics, Vol. 2. Cambridge University Press. (pp. 219-229)de:Catalan-Zahl

ko:카탈란 수 it:Numero di Catalan he:מספר קטלן ru:Число Kaталана sl:Catalanovo število


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