In combinatorial mathematics, the Catalan numbers form a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects. They are named after the Belgian mathematician Eugène Charles Catalan (1814–1894). The nth Catalan number is given directly in terms of binomial coefficients by \( C_n = \frac{1}{n+1}{2n\choose n} = \frac{(2n)!}{(n+1)!\,n!} = \prod\limits_{k=2}^{n}\frac{n+k}{k} \qquad\mbox{ for }n\ge 0. \) The first Catalan numbers 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, … (sequence A000108 in OEIS)
Properties An alternative expression for Cn is \( C_n = {2n\choose n}  {2n\choose n+1} \quad\text{ for }n\ge 0, \) which is equivalent to the expression given above because \( \tbinom{2n}{n+1}=\tfrac{n}{n+1}\tbinom{2n}n \). This shows that Cn is an integer number, 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 \( C_0 = 1 \quad \mbox{and} \quad C_{n+1}=\sum_{i=0}^{n}C_i\,C_{ni}\quad\text{for }n\ge 0; \) moreover, \( C_n= \frac 1{n+1} \sum_{i=0}^n {n \choose i}^2. \) This is due to the fact that {2n\choose n} = \sum_{i=0}^n {n \choose 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 ni numbers from the remaining n numbers. They also satisfy: \( C_0 = 1 \quad \mbox{and} \quad C_{n+1}=\frac{2(2n+1)}{n+2}C_n, \) which can be a more efficient way to calculate them. Asymptotically, the Catalan numbers grow as \( C_n \sim \frac{4^n}{n^{3/2}\sqrt{\pi}} \) in the sense that the quotient of the nth Catalan number and the expression on the right tends towards 1 as n → +∞. (This can be proved by using Stirling's approximation for n!.) The only Catalan numbers Cn that are odd are those for which n = 2k − 1. All others are even. Catalan numbers have integral represantation \( C_n=\int_0^4x^n\rho(x)dx \) where \rho(x)=\frac 1{2\pi}\sqrt{\frac{4x}{x}}. It's mean that Catalan numbers are solution of generalized Hausdorff moment problem. Orthogonal polynomials with weight \rho(x) on [0;4] have a form \( H_n(x)=\sum_{k=0}^n{n+k \choose nk}(x)^k. \) 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. 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 (see also Dyck language). For example, the following are the Dyck words of length 6: XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY. Reinterpreting 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 (or the number of ways of associating n applications of a binary operator). For n = 3, for example, we have the following five different parenthesizations of four factors: \( ((ab)c)d \quad (a(bc))d \quad(ab)(cd) \quad a((bc)d) \quad a(b(cd)) \) 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 Cn is the number of full binary trees with n + 1 leaves: Catalan number binary tree example.png If the leaves are labelled, we have the quadruple factorial numbers. Cn is the number of nonisomorphic ordered trees with n+1 vertices. (An ordered tree is a rooted tree in which the children of each vertex are given a fixed lefttoright order.)[1] Cn is the number of monotonic paths 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: Catalan number 4x4 grid example 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 hexagons illustrate the case n = 4: CatalanHexagonsexample C_{n} is the number of stacksortable permutations of {1, ..., n}. A permutation w is called stacksortable 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 oneelement sequences. These are the permutations that avoid the pattern 231. C_{n} is the number of permutations of {1, ..., n} that avoid the pattern 123 (or any of the other patterns of length 3); that is, the number of permutations with no threeterm 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_{n} is the number of noncrossing partitions of the set {1, ..., n}. A fortiori, C_{n} never exceeds the nth Bell number. C_{n} is also the number of noncrossing partitions of the set {1, ..., 2n} 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 cumulants of degree more than 2 of the Wigner semicircle law are zero. This law is important in free probability theory and the theory of random matrices. C_{n} is the number of ways to tile a stairstep shape of height n with n rectangles. The following figure illustrates the case n = 4: Catalan stairsteps 4.svg C_{n} is the number of standard Young tableaux whose diagram is a 2byn rectangle. In other words, it is the number ways the numbers 1, 2, ..., 2n can be arranged in a 2byn rectangle so that each row and each column is increasing. As such, the formula can be derived as a special case of the hooklength formula. C_{n} is the number of ways that the vertices of a convex 2ngon can be paired so that the line segments joining paired vertices do not intersect. C_{n} is the number of semiorders on n unlabeled items.[2] Proof of the formula There are several ways of explaining why the formula \( C_n = \frac{1}{n+1}{2n\choose n} \) 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[3] recurrence relation \( C_0 = 1 \quad \text{and} \quad C_{n+1}=\sum_{i=0}^n C_i\,C_{ni}\quad\text{for }n\ge 0. \) 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 \( c(x)=\sum_{n=0}^\infty C_n x^n. \) The two recurrence relations together can then be summarized in generating function form by the relation \( c(x)=1+xc(x)^2;\, \) in other words, this equation follows from the recurrence relations by expanding both sides into power series. On the one hand, the recurrence relations uniquely determine the Catalan numbers; on the other hand, the generating function solution \( c(x) = \frac{1\sqrt{14x}}{2x}=\frac{2}{1+\sqrt{14x}} \) has a power series at 0 and its coefficients must therefore be the Catalan numbers. (Since the other solution has a pole at 0, this reasoning doesn't apply to it.) The square root term can be expanded as a power series using the identity \( \sqrt{1+y} = \sum_{n=0}^\infty {\frac12 \choose n} y^n = 1  2\sum_{n=1}^\infty {2n2 \choose n1} \left(\frac{1}{4}\right)^n \frac{y^n}{n}. \) 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 \( c(x) = \sum_{n=0}^\infty {2n \choose n} \frac{x^n}{n+1}. \) The coefficients are now the desired formula for Cn. Another way to get c(x) is to solve for xc(x) and observe that \( \int_0^x \! t^n \, dt \) appears in each term of the power series. This proof depends on a trick known as André's reflection method (not to be confused with the Schwarz reflection principle in complex analysis), 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.[4] It is most easily expressed in terms of the "monotonic paths which do not cross the diagonal" problem (see above). 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 \( {2n\choose n+1}. \) 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 \( {2n\choose n}{2n\choose n+1} \) which is the nth Catalan number Cn. 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. 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. 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 \( {2n\choose n} \) monotonic paths, we obtain the desired formula \( C_n = \frac{1}{n+1}{2n\choose n}. \) Figure 4. All monotonic paths in a 3×3 grid, illustrating the exceedancedecreasing algorithm. 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 Cn and Cn+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)Cn 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)Cn+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 \( (4n+2)C_n = (n+2)C_{n+1}. \) The binomial formula for Cn 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 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 \( C_0 = 1 \quad \text{and} \quad C_{n+1} = \sum_{i=0}^n C_i\,C_{ni}\quad\text{for }n\ge 0. \) Now let b stand for a balanced string of length 2n containing an equal number of "[" and "]" and \textstyle B_n = {2n\choose n} = d_n C_n with some factor dn ≥ 1. As above, any balanced string can be uniquely decomposed into either [ c ] b or ] c+[b, so \( B_{n+1} = 2 \sum_{i=0}^n B_i C_{ni}. \) Also, any incorrect balanced string starts with c ], so B_{n+1}  C_{n+1} = \sum_{i=0}^n {2i+1 \choose i} C_{ni} = \sum_{i=0}^n \frac{2i+1}{i+1} B_i C_{ni}. Subtracting the above equations and using Bi = di Ci gives \( C_{n+1} = 2 \sum_{i=0}^n d_i C_i C_{ni}  \sum_{i=0}^n \frac{2i+1}{i+1} d_i C_i C_{ni} = \sum_{i=0}^n \frac{d_i}{i+1} C_i C_{ni}. \) Comparing coefficients with the original recursion formula for Cn gives di = i + 1, so \( C_n = \frac{1}{n+1}{2n\choose n}. \) Hankel matrix 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 \( \det\begin{bmatrix}1 & 1 & 2 & 5 \\ 1 & 2 & 5 & 14 \\ 2 & 5 & 14 & 42 \\ 5 & 14 & 42 & 132\end{bmatrix} = 1. \) Note that if the entries are "shifted", namely the Catalan numbers Ci+j−1, the determinant is still 1, regardless of the size of n. For example, for n = 4 we have \( \det\begin{bmatrix}1 & 2 & 5 & 14 \\ 2 & 5 & 14 & 42 \\ 5 & 14 & 42 & 132 \\ 14 & 42 & 132 & 429 \end{bmatrix} = 1. \) The Catalan numbers form the unique sequence with this property.[citation needed] The quadruple factorial is given by \frac{(2n)!}{n!}, or \left(n+1\right)! C_n. This is the solution to labelled variants of the above combinatorics problems. It is entirely distinct from the multifactorials. 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 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 in the Chinese publication Neimenggu Daxue Xuebao that the Catalan number sequence had been used in China by the mathematician Antu Ming 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 Antu Ming, 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). Associahedron
Catalan's problem Lobb numbers Notes ^ Stanley p.221 example (e) References Conway and Guy (1996) The Book of Numbers. New York: Copernicus, pp. 96–106. Retrieved from "http://en.wikipedia.org/"


