Catalan number

We first observe that all of the combinatorial problems listed above satisfy Segner's[9] recurrence relation

${\displaystyle C_{0}=1\quad {\text{and}}\quad C_{n+1}=\sum _{i=0}^{n}C_{i}\,C_{n-i}\quad {\text{for }}n\geq 0.}$

For example, 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

${\displaystyle c(x)=\sum _{n=0}^{\infty }C_{n}x^{n}.}$

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

${\displaystyle c(x)=1+xc(x)^{2};}$

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, the generating function relation can be algebraically solved to yield

${\displaystyle c(x)={\frac {1\mp {\sqrt {1-4x}}}{2x}}={\frac {2}{1\pm {\sqrt {1-4x}}}}.}$

Choosing the minus sign (in the first expression), the fraction has a power series at 0 so its coefficients must therefore be the Catalan numbers. This solution satisfies

${\displaystyle \lim _{x\to 0^{+}}c(x)=C_{0}=1}$

The other solution, with the plus sign, has a pole at 0 so it cannot be a valid solution for c(x).

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

${\displaystyle {\sqrt {1+y}}=\sum _{n=0}^{\infty }{{\frac {1}{2}} \choose n}y^{n}=\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}}{4^{n}(2n-1)}}{2n \choose n}y^{n}=1+{\frac {1}{2}}y-{\frac {1}{8}}y^{2}+\cdots .}$

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

${\displaystyle c(x)=\sum _{n=0}^{\infty }{2n \choose n}{\frac {x^{n}}{n+1}}.}$

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 ${\displaystyle \int _{0}^{x}\!t^{n}\,dt}$ appears in each term of the power series.

Figure 1. The invalid portion of the path (in solid red color) is flipped. Bad paths reach (n – 1, n + 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.[10]) 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 ${\displaystyle {\tbinom {2n}{n}}}$ 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,

${\displaystyle {n-1+n+1 \choose n-1}={2n \choose n-1}={2n \choose n+1}}$

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,

${\displaystyle C_{n}={2n \choose n}-{2n \choose n+1}.}$

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.

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).[11]

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.

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.

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 − 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

${\displaystyle {2n \choose n}}$

monotonic paths, we obtain the desired formula

${\displaystyle C_{n}={\frac {1}{n+1}}{2n \choose n}.}$

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.

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

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

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

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

${\displaystyle C_{0}=1\quad {\text{and}}\quad C_{n+1}=\sum _{i=0}^{n}C_{i}\,C_{n-i}\quad {\text{for }}n\geq 0.}$

Now let b stand for a balanced string of length 2n—that is, containing an equal number of "[" and "]"—and ${\displaystyle \textstyle B_{n}={2n \choose n}=d_{n}C_{n}}$ with some factor d_n ≥ 1. As above, any balanced string can be uniquely decomposed into either [ c ] b or ] c⁺ [ b, so

${\displaystyle B_{n+1}=2\sum _{i=0}^{n}B_{i}C_{n-i}.}$

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

${\displaystyle B_{n+1}-C_{n+1}=\sum _{i=0}^{n}{2i+1 \choose i}C_{n-i}=\sum _{i=0}^{n}{\frac {2i+1}{i+1}}B_{i}C_{n-i}.}$

Subtracting the above equations and using B_i = d_i C_i gives

${\displaystyle C_{n+1}=2\sum _{i=0}^{n}d_{i}C_{i}C_{n-i}-\sum _{i=0}^{n}{\frac {2i+1}{i+1}}d_{i}C_{i}C_{n-i}=\sum _{i=0}^{n}{\frac {d_{i}}{i+1}}C_{i}C_{n-i}.}$

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

${\displaystyle C_{n}={\frac {1}{n+1}}{2n \choose n}.}$

This simple proof[12] is also based on the Dyck words interpretation of the Catalan numbers but uses the beautiful Cycle Lemma of Dvoretzky and Motzkin.[13] 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 ${\displaystyle m}$ X's and ${\displaystyle n}$ Y's, where ${\displaystyle m>n}$, has precisely ${\displaystyle m-n}$ dominating cyclic permutations. To see this, just arrange the given sequence of ${\displaystyle m+n}$ X's and Y's in a circle and repeatedly remove adjacent pairs XY until only ${\displaystyle m-n}$ X's remain. Each of these X's was the start of a dominating cyclic permutation before anything was removed. In particular, when ${\displaystyle m=n+1}$, 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 ${\displaystyle C_{n}={\frac {1}{2n+1}}{2n+1 \choose n}={\frac {1}{n+1}}{2n \choose n}}$ distinct cycles of ${\displaystyle n+1}$ X's and ${\displaystyle n}$ Y's, each of which corresponds to exactly one Dyck sequence, ${\displaystyle C_{n}}$ counts Dyck sequences.