加泰罗尼亚数(Catalan numbers)
加泰罗尼亚的数字形成的自然数,在不同的计数问题经常发生涉及递归定义的对象,序列.它们被命名比利时数学家欧仁查理加泰罗尼亚（1814年至1894年）.
设C(m,n)=m!/[(m-n)!n!]
第n个Catalan数,是直接二项式系数项
c(2n,n)/(n+1)
对于n=0,1,2,3,.
加泰罗尼亚数: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,
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(2n,n)/(n+1)
The first Catalan numbers (sequence A000108 in OEIS) 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,
这是排列组合中的组合公式
C(2n,n)=(2n)!/[(2n-n)!n!]
从2n项中选择n项的组合方式的数量
n=3时,是从2n=2*3=6项中选择n=3项的组合方式数量
C(3,6)=6!/(3!(6-3)!=(6*5*4*3*2)/{(3*2)(3*2)]=20
