FIG Ming (Cattleya) Number

I do not know the source of reprint a blog summary

Catalan number is combinatorics a count often appear in various problems occurring in the number of columns . It made to the Belgian mathematician Eugène Charles Catalan ( 1814 - 1894 named).

General Formula of Catalan number of $C_n = \frac{1}{n+1}{2n \choose n} = \frac{(2n)!}{(n+1)!n!}$ alternative recursion formula: h (n) = (( 4 * n-2) / (n + 1)) * h (n-1);

The first few of ( OEIS number column A000108 ):. 1,. 1, 2,. 5, 14, 42 is, 132, 429, 1430, 4862, 16796, 58786, 208 012, 742.9 thousand, 2.67444 million, 9694845, 35.35767 million, 129.64479 million, 477 638 700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, ...

nature

C _n- other forms of expression is $C_n = {2n\choose n} - {2n\choose n-1} \quad\mbox{ for }n\ge 1$ therefore, C _n- is a natural number ; this is not apparent from the general formula of the previous pass. This form of expression is the basis of the previous formula André proof. (See below second proof .)

Catalan number satisfy the following recurrence relations

$C_0 = 1 \quad \mbox{and} \quad C_{n+1}=\sum_{i=0}^{n}C_i\,C_{n-i}\quad\mbox{for }n\ge 0.$

It also satisfies

$C_0 = 1 \quad \mbox{and} \quad C_{n+1}=\frac{2(2n+1)}{n+2}C_n,$

This provides a faster method to calculate the Catalan number.

Asymptotic growth of the number of Catalan

$C_n \sim \frac{4^n}{n^{3/2}\sqrt{\pi}}$

It is the meaning of the left and right divided type commercially formula tends to 1 when the n- → ∞. (This may be the n- ! Of Stirling's approximation to prove it.)

All chica taran number C _n- satisfy $n- = 2 K -. 1. All other numbers are even Catalan.$

application

Combinatorial Mathematics has very much. The combined structure may be used to count the number of Catalan. Enumerative Combinatorics in Richard P. Stanley: The Problem in Volume 2 of a book structure comprising a composition expressed by 66 distinct numbers of Catalan. Using the following C _n- =. 3 and C _n- = Example 4 For a number of:

C _n- represents the length of 2n number of dyck word. Dyck word is an n X's and n strings consisting of Y, and the partial strings are all satisfying number equal to Y is greater than the number of X's. The following is a length of dyck words 6:

XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY

The above example X into left parenthesis, Y replaced by right parenthesis, C _n represents all comprising n valid set of parentheses the number of expressions of:

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

C _n- expressed n + 1 th leaves binary number.

C _n represents a different configuration containing all n full branchings node binary number. (A root of the binary tree is full if and only if every node has two sub-tree or subtree not.)

prove:

1 shows make the stack, the stack shows 0, can be converted into a request 2n bits, containing n number 1, n when a binary 0, to meet any of a scanning from left to right, through the small number 0 1 in number. Obviously having n number. 1, n zeros 2n -bit binary number total ${2n \choose n}$ number, the number of the following considerations not meet the requirements.

Consider having a n a 1, n a 2n-bit binary number 0, the first scan + 1 2m when the bit m + 1 th 0 and m number 1 (easy to prove that there is a certain case), then the backward 0- 1, the arrangement must nm ones and nm-1 zeros. The 2m + 2 section becomes 0 after 1,1 and becomes 0, then the corresponding one of n + 1 th 0 and n-1 binary 1s. Vice versa (both similar ideas correspond proof).

Thereby $C_n = {2n \choose n} - {2n \choose n + 1} = \frac{1}{n+1}{2n \choose n}$ . QED.

C _n- represents all n- × n- not cross diagonal lattice points monotonically path number. A monotone path from the lower left corner of the grid points, ending at the top right of the grid points, each step are up or to the right. Counting the number of such paths is equivalent to the number calculated Dyck word: X stands for "right", Y stands for "up." The following figure shows n- = 4 in the case of:

C _n- represents the apex by connecting n- + 2 side of the convex polygon is divided into triangles number of methods. The figure is a n- = 4 in the case of:

C_n表示对{1, ..., n}依序进出栈的置换个数。一个置换w是依序进出栈的当S(w) = (1, ..., n), 其中S(w)递归定义如下：令w = unv，其中n为w的最大元素，u和v为更短的数列；再令S(w) =S(u)S(v)n，其中S为所有含一个元素的数列的单位元。

C_n表示集合{1, ..., n}的不交叉划分的个数. 那么, C_n 永远不大于第n项贝尔数. C_n也表示集合{1, ..., 2n}的不交叉划分的个数，其中每个段落的长度为2。综合这两个结论，可以用数学归纳法证明 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表示用n个长方形填充一个高度为n的阶梯状图形的方法个数。下图为 n = 4的情况:

百度百科资料:
简介

　　中文:卡特兰数
　　Catalan数是组合数学中一个常出现在各种计数问题中出现的数列。由以比利时的数学家欧仁·查理·卡塔兰 (1814–1894)命名。
　　原理：
　　令h(0)=1,h(1)＝1,catalan数满足递归式：
　　h(n)= h(0)*h(n-1) + h(1)*h(n-2) + + h(n-1)h(0) (其中n>=2)
　　该递推关系的解为：
　　h(n)=C(2n,n)/(n + 1) (n=1,2,3,)
另类递归式： h(n)=((4*n-2)/(n+1))*h(n-1);
　　
　　前几项为（OEIS中的数列A000108）: 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,
应用

　　我总结了一下，最典型的四类应用：（实质上却都一样，无非是递归等式的应用，就看你能不能分解问题写出递归式了）
1.括号化问题。

　　矩阵链乘： P=a1×a2×a3×……×an，依据乘法结合律，不改变其顺序，只用括号表示成对的乘积，试问有几种括号化的方案？(h(n)种)
2. Stack the order of questions.

　　A stack (infinity) into the stack of sequence 1,2,3, .. n, the number of different pop sequence?
　　Something like:
　　(1) there are 2n personal lined up to enter the theater. Admission $ 5. Of which only n individuals have a five dollar bill, while only 10 people n dollar bills, no other theater bills, asked how many methods so that as long as people buy tickets 10 yuan, ticket office there is a $ 5 bill change? (5 yuan by the holder reaches regarded as the stack 5 yuan, holding 10 yuan considered that the person reaches the stack a stack 5 yuan)
　　(2) select 2n points on a circle, connecting these points together in pairs, the number n of line segments so that the resulting disjoint method.
3. The line is divided into triangles multilateral issues.

　　? The number of ways a convex polygon area is divided into triangular areas
　　similar: a big city lawyer in n n blocks east and one block north of her home. She walked 2n blocks every day to go to work. If she
　　never crossing (but can be touch) diagonal from home to the office, then how many possible way?
　　Similar: 2n selected points on the circle, so that the number of ways to connect the n line segments obtained disjoint pairs of these points?
4 to the top node in the composition of the binary tree.

　　Given N nodes, how many different shapes can constitute a binary tree?
　　(Binary tree must be!
　　Go to a point as a vertex, then the left turn can be 0 to the N-1 corresponding to the N-1 to the right is zero, match the two multiplied, that h (0) * h ( . 1-n-) + H (2) * H (2-n-) + + H (. 1-n-) H (0) = H (n-))
　　(can form h (N) th)