Catalan number
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 "n"th Catalan number can be expressed directly in terms of binomial coefficients by The first Catalan numbers for "n" = 0, 1, 2, 3, ... are Properties. An alternative expression for "C" is which is equivalent to the expression given above because formula_4. This expression shows that "C" 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 and Asymptotically, the Catalan numbers grow as formula_7 in the sense that the quotient of the "n"th Catalan number and the expression on the right tends towards 1 as "n" approaches infinity. This can be proved by using the asymptotic growth of the central binomial coefficients, by Stirling's approximation for formula_8, or via generating functions. The only Catalan numbers "C" that are odd are those for which "n" = 2 − 1; all others are even. The only prime Catalan  numbers are "C" = 2 and "C" = 5. The Catalan numbers have the integral representations The latter representation is closely connected to Wigner's semicircle law for eigenvalue distribution of random symmetric matrices. 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 "C" = 5 and "C" = 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: 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 "w" and "w". 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 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" = −4"x" gives and substituting this power series into the expression for  c("x"), the expansion simplifies to Let formula_29, so that and because formula_31 (see 'proof of recurrence' above) we have Second proof. 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 2"n" steps are up or right, there are in total formula_33 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 2"n" 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. Third proof. This bijective proof provides a natural explanation for the term "n" + 1 appearing in the denominator of the formula for "C". 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 formula_36 monotonic paths, we obtain the desired formula formula_37 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, "C" = 5, and the last column displays all paths no higher than the diagonal. Using Dyck words, start with a sequence from formula_38. Let formula_39 be the first that brings an initial subsequence to equality, and configure the sequence as formula_40. The new sequence is formula_41. Fourth proof. This proof uses the triangulation definition of Catalan numbers to establish a relation between "C" and "C". 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 2"n" + 1 total edges. There are (4"n" + 2)"C" 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)"C" 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. Thus Write formula_43 Because formula_44 then Applying the recursion with formula_46 gives the result. Fifth proof. This proof is based on the Dyck words interpretation of the Catalan numbers, so "C" 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" = [ "c" ] "c", summing over the possible spots to place the closing bracket immediately gives the recursive definition Let "b" stand for a "balanced" string of length 2"n"—that is, containing an equal number of "[" and "]", so formula_48. 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: Therefore Sixth proof. 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 formula_55 X's and formula_56 Y's, where formula_57, has precisely formula_58 dominating cyclic permutations. To see this, arrange the given sequence of formula_59 X's and Y's in a circle. Repeatedly removing XY pairs leaves exactly formula_58 X's. Each of these X's was the start of a dominating cyclic permutation before anything  was removed. For example, consider formula_61. This is dominating, but none of its cyclic permutations formula_62, formula_63, formula_64 and formula_65 are. In particular, when formula_66, 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 formula_67 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 formula_68 distinct cycles of formula_69 X's and formula_56 Y's, each of which corresponds to exactly one Dyck sequence, hence formula_71 counts Dyck  sequences. Hankel matrix. The "n"×"n" Hankel matrix whose ("i", "j") entry is the Catalan number "C" 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" 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. et cetera. 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 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 sin(2α) and sin(4α) in terms of sin(α). Generalizations . The two-parameter sequence of non-negative integers formula_78 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 formula_79, this is just two times the ordinary Catalan numbers, and for formula_80, the numbers have an easy combinatorial description. However, other combinatorial descriptions are only known for formula_81 and formula_82, 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 formula_10 corresponds to the root system of type formula_84. 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. Catalan number In