Catalan number

Properties

An alternative expression for C_n is

which is equivalent to the expression given above because ( 2 n n + 1 ) = n n + 1 ( 2 n n ) . This 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.

The Catalan numbers satisfy the recurrence relation

moreover,

This is because ( 2 n n ) = ∑ i = 0 n ( n i ) 2 , since choosing n numbers from a 2n set of numbers can be uniquely divided into 2 parts: choosing i numbers out of the first n numbers and then choosing n-i numbers from the remaining n numbers.

They also satisfy:

which can be a more efficient way to calculate them.

Asymptotically, the Catalan numbers grow as

in the sense that the quotient of the nth Catalan number and the expression on the right tends towards 1 as n → +∞. Some sources use just C n ≈ 4 n n 3 / 2 . (This can be proved by using Stirling's approximation for n!.)

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.

The Catalan numbers have an integral representation

where ρ ( x ) = 1 2 π 4 − x x . 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 ρ ( x ) on [ 0 , 4 ] are

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.

The following diagrams show the case n = 4:

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

Proof of the formula

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.

First proof

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, the generating function relation can be algebraically solved to yield

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

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

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

The coefficients are now the desired formula for C_n.

Another way to get c(x) is to solve for xc(x) and observe that ∫ 0 x t n d t appears in each term of the power series.

Second proof

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.) We count the paths which start and end on the diagonal of the n × n grid. All such paths have n rightward and n upward steps. Since we can choose which of the 2n steps are upward (or, equivalently, rightward) ones, there are ( 2 n n ) 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 2n steps, there must now be n + 1 upward steps and n - 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 - 1, n + 1). As any monotonic path in the n - 1 × n + 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,

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. At that first Y, there are k + 1 Y's and k X's for some k between 1 and n - 1.

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 C_n. 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.

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.

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.

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, 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.

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 − 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.

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 P with n+ 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 2n+1 edges. There are (4n+2)C_n such decorated triangulations. Now given a polygon Q with n+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+2)C_n+1 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

The binomial formula for C_n follows immediately from this relation and the initial condition C₁ = 1.

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 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

Now let b stand for a balanced string of length 2n—that is, containing an equal number of "[" and "]"—and B n = ( 2 n n ) = d n C n with some factor d_n ≥ 1. As above, any balanced string can be uniquely decomposed into either [ c ] b or ] c⁺ [ b, so

Also, any incorrect balanced string starts with c ], so

Subtracting the above equations and using B_i = d_i C_i gives

Comparing coefficients with the original recursion formula for C_n gives d_i = i + 1, so

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 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 C_i+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.

History

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. That is when he started to write his book Ge Yuan Mi Lu Jie Fa, 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 ( 2 m ) ! ( 2 n ) ! ( 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 and m = 3 , and it is an open problem to find a general combinatorial interpretation.