Catalan number

Properties

An alternative expression for C_n is

$C_{n}={2n \choose n}-{2n \choose n+1}\quad {\text{ for }}n\geq 0,$

which is equivalent to the expression given above because ${\tbinom {2n}{n+1}}={\tfrac {n}{n+1}}{\tbinom {2n}{n}}$. 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

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

moreover,

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

This is due to the fact that ${\tbinom {2n}{n}}=\sum _{i=0}^{n}{\tbinom {n}{i}}^{2},$ 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:

$C_{0}=1\quad {\mbox{and}}\quad C_{n+1}={\frac {2(2n+1)}{n+2}}C_{n},$

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 C_n that are odd are those for which n = 2^k − 1. All others are even.

The Catalan numbers have an integral representation

$C_{n}=\int _{0}^{4}x^{n}\rho (x)dx$

where $\rho (x)={\tfrac {1}{2\pi }}{\sqrt {\tfrac {4-x}{x}}}.$ 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 $\rho (x)$ on $[0,4]$ are

$H_{n}(x)=\sum _{k=0}^{n}{n+k \choose n-k}(-x)^{k}.$

Applications in combinatorics

The associahedron of order 4 with C₄=14 vertices

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.

• C_n 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.

• Re-interpreting the symbol X as an open parenthesis and Y as a close parenthesis, C_n counts the number of expressions containing n pairs of parentheses which are correctly matched:

((()))     ()(())     ()()()     (())()     (()())

• C_n 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 C_n is the number of full binary trees with n + 1 leaves:

If the leaves are labelled, we have the quadruple factorial numbers.

• C_n is the number of non-isomorphic ordered trees with n+1 vertices. (An ordered tree is a rooted tree in which the children of each vertex are given a fixed left-to-right order.)

• C_n 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:

A succinct representation is listing the Catalan elements by column height:

[0,0,0,0][0,0,0,1][0,0,0,2][0,0,1,1]

[0,1,1,1] [0,0,1,2] [0,0,0,3] [0,1,1,2][0,0,2,2][0,0,1,3]

[0,0,2,3][0,1,1,3] [0,1,2,2][0,1,2,3]

• C_n 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:

• C_n is the number of stack-sortable permutations of {1, ..., n}. A permutation w is called stack-sortable 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 one-element 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 three-term 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:

• C_n is the number of standard Young tableaux whose diagram is a 2-by-n rectangle. In other words, it is the number of ways the numbers 1, 2, ..., 2n can be arranged in a 2-by-n rectangle so that each row and each column is increasing. As such, the formula can be derived as a special case of the hook-length formula.

• C_n is the number of ways that the vertices of a convex 2n-gon can be paired so that the line segments joining paired vertices do not intersect. This is precisely the condition that guarantees that the paired edges can be identified (sewn together) to form a closed surface of genus zero (a topological 2-sphere).

• C_n is the number of semiorders on n unlabeled items.

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.

First proof

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

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

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

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

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

$c(x)={\frac {1-{\sqrt {1-4x}}}{2x}}={\frac {2}{1+{\sqrt {1-4x}}}}$

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

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

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 C_n.

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.

Second proof

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. It is most easily expressed in terms of the "monotonic paths which do not cross the diagonal" problem (see above).

File:Catalan number reflection example.png

Figure 1. The green portion of the path is flipped.

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 C_n.

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).

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.

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.

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

${2n \choose n}$

monotonic paths, we obtain the desired formula

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

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.

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

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

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

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

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

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

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

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

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

Subtracting the above equations and using B_i = d_i C_i gives

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

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

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

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

$\det {\begin{bmatrix}1&1&2&5\\1&2&5&14\\2&5&14&42\\5&14&42&132\end{bmatrix}}=1.$

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

$\det {\begin{bmatrix}1&2&5&14\\2&5&14&42\\5&14&42&132\\14&42&132&429\end{bmatrix}}=1.$

Taken together, these two conditions uniquely define the Catalan numbers.

Quadruple factorial

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.

History

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 an Inner Mongolia University of Technology publication that the Catalan number sequence had been used in China by the Mongolian mathematician Minggantu 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 Minggantu, 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).

See also

• Associahedron
• Bertrand's ballot theorem
• Binomial transform
• Partition related number triangles

• Catalan's problem
• Catalan–Mersenne number
• List of factorial and binomial topics

• Lobb numbers
• Narayana number
• Tamari lattice

References

• Conway and Guy (1996) The Book of Numbers. New York: Copernicus, pp. 96–106.
• Gardner, Martin (1988), Time Travel and Other Mathematical Bewilderments, New York: W.H. Freeman and Company, pp. 253–266 (Ch. 20), ISBN 0-7167-1924-X
• Koshy, Thomas (2008), Catalan Numbers with Applications, Oxford University Press, ISBN 0-19-533454-X
• Koshy, Thomas & Zhenguang Gao (2011) "Some divisibility properties of Catalan numbers", Mathematical Gazette 95:96–102.
• Larcombe, P.J. (1999) "The 18th century Chinese discovery of the Catalan numbers", Mathematical Spectrum 32:5–7.
• Stanley, Richard P. (1999), Enumerative combinatorics. Vol. 2, Cambridge Studies in Advanced Mathematics, vol. 62, Cambridge University Press, ISBN 978-0-521-56069-6; 978-0-521-78987-5, MR 1676282 {{citation}}: Check |isbn= value: invalid character (help)
• Egecioglu, Omer (2009), A Catalan-Hankel Determinant Evaluation (PDF)

External links

• Stanley, Richard P. (1998), Catalan addendum to Enumerative Combinatorics, Volume 2 (PDF)
• Weisstein, Eric W. "Catalan Number". MathWorld.
• Dickau, Robert M.: Catalan numbers Further examples.
• Davis, Tom: Catalan numbers. Still more examples.
• Schmidthammer, Jürgen: Catalan-Zahlen Zulassungsarbeit zum Staatsexamen (PDF-File; 7,05 MB)
• "Equivalence of Three Catalan Number Interpretations" from The Wolfram Demonstrations Project [1]