f n = f 0 f n − 1 + f 1 f n − 2 + . . . + f n − 1 f 0 f_n=f_0f_{n-1}+f_1f_{n-2}+...+f_{n-1}f_0 fn=f0fn−1+f1fn−2+...+fn−1f0
也即
f n = ∑ i = 0 n − 1 f i f n − 1 − i f_n=\sum\limits_{i=0}^{n-1} f_i f_{n-1-i} fn=i=0∑n−1fifn−1−i
卡特兰一般在计数题中有运用
一般采用组合数的计算方式
由折线法可知
f n = C 2 n n − C 2 n n + 1 f_n=C^n_{2n}-C^{n+1}_{2n} fn=C2nn−C2nn+1
= ( 2 n ) ! n ! ( 2 n − n ) ! − ( 2 n ) ! ( n + 1 ) ! ( 2 n − ( n + 1 ) ) ! =\frac{(2n)!}{n!(2n-n)!}-\frac{(2n)!}{(n+1)!(2n-(n+1))!} =n!(2n−n)!(2n)!−(n+1)!(2n−(n+1))!(2n)!
= ( 2 n ) ! ( n ! ) 2 − ( 2 n ) ! ( n + 1 ) ! ( n − 1 ) ! =\frac{(2n)!}{(n!)^2}-\frac{(2n)!}{(n+1)!(n-1)!} =(n!)2(2n)!−(n+1)!(n−1)!(2n)!
= ( 2 n ) ! ( n ! ) 2 − ( 2 n ) ! ( n ) ! 2 n n + 1 =\frac{(2n)!}{(n!)^2}-\frac{(2n)!}{(n)!^2}\frac{n}{n+1} =(n!)2(2n)!−(n)!2(2n)!n+1n
= 1 n + 1 ( 2 n ) ! ( n ) ! 2 =\frac{1}{n+1}\frac{(2n)!}{(n)!^2} =n+11(n)!2(2n)!
= ∏ i = n + 2 2 n i ( n ) ! =\frac{\prod\limits^{2n}_{i=n+2}i}{(n)!} =(n)!i=n+2∏2ni
f n + 1 = 4 n + 2 n + 2 f n f_{n+1}=\frac{4n+2}{n+2}f_n fn+1=n+24n+2fn
做题的时候先用定义法证一下,实在不行就打表猜结论好啦QAQ