In combinatorial mathematics, the Catalan numbers are a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects. They are named after the French-Belgian mathematician Eugène Charles Catalan (1814–1894).
The nth Catalan number can be expressed directly in terms of binomial coefficients by
The first Catalan numbers for n = 0, 1, 2, 3, ... are
An alternative expression for Cn is
which is equivalent to the expression given above because . This expression shows that Cn 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.
The Catalan numbers satisfy the recurrence relations
Asymptotically, the Catalan numbers grow as
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 the integral representations
which immediately yields .
The latter representation is closely connected to Wigner's semicircle law for eigenvalue distribution of random symmetric matrices.
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 is . Since the 1D random walk is recurrent, the probability that the walker eventually arrives at -1 is .
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.
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.
The following diagrams show the case n = 4:
This can be represented by listing the Catalan elements by column height:
/\/\, / \
/\ /\ /\/\ / \
/\/\/\, /\/ \, / \/\, / \, / \
There are several ways of explaining why the formula
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.
We first observe that all of the combinatorial problems listed above satisfy Segner's recurrence relation
For example, every Dyck word w of length ≥ 2 can be written in a unique way in the form
with (possibly empty) Dyck words w1 and w2.
The generating function for the Catalan numbers is defined by
The recurrence relation given above can then be summarized in generating function form by the relation
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 c and using the quadratic formula, the generating function relation can be algebraically solved to yield two solution possibilities
From the two possibilities, the second must be chosen because only the choice of the second gives
The square root term can be expanded as a power series using the identity
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
and substituting this power series into the expression for c(x), the expansion simplifies to
Let , so that
and because (see 'proof of recurrence' above)
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 monotonic paths of this type. A bad path crosses the main diagonal and touches the next higher diagonal (red in the illustration).
The part of the path after the higher diagonal is flipped about that diagonal, as illustrated with the red 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 higher 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:
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,
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 Xs.
This bijective proof provides a 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).
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.
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 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 unique vertical edge that passes from above the diagonal to below it - all the other vertical edges stay on the same side of the diagonal.
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 monotonic paths, we obtain the desired formula
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 . Let be the first X that brings an initial subsequence to equality, and configure the sequence as . The new sequence is .
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.
Applying the recursion with gives the result.
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
Let b stand for a balanced string of length 2n—that is, containing an equal number of "[" and "]", so . As before, any balanced string can be uniquely decomposed into either [ c ] b or ] c' [ b, so
Any incorrect balanced string starts with c ], and the remaining string has one more [ than ], so
Also, from the definitions, we have:
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 X's and Y's, where , has precisely dominating cyclic permutations. To see this, arrange the given sequence of X's and Y's in a circle. Repeatedly removing XY pairs leaves exactly X's. Each of these X's was the start of a dominating cyclic permutation before anything was removed.
For example, consider . This is dominating, but none of its cyclic permutations , , and are.
In particular, when , 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 in total, and each one belongs to an equivalence class of size 2n+1 (because n, m and 2n+1 are pairwise coprime), we have distinct cycles of X's and Y's, each of which corresponds to exactly one Dyck sequence, hence counts Dyck sequences.
The n×n Hankel matrix whose (i, j) entry is the Catalan number Ci+j−2 has determinant 1, regardless of the value of n. For example, for n = 4 we have
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
Taken together, these two conditions uniquely define the Catalan numbers.
Another feature unique to the Catalan–Hankel matrix is the determinant of the n×n submatrix starting at 2 has determinant n + 1.
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 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 and in terms of .
The Catalan numbers can be interpreted as a special case of the Bertrand's ballot theorem. Specifically, is the number of ways for a candidate A with n+1 votes to lead candidate B with n votes.
The two-parameter sequence of non-negative integers 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 , this is just two times the ordinary Catalan numbers, and for , the numbers have an easy combinatorial description. However, other combinatorial descriptions are only known for and , 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 corresponds to the root system of type . 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 k-fold convolution is:
The Catalan numbers are a solution of a version of the Hausdorff moment problem.