Catalan数表达式完整推导

写在前面

推导一下Catalan数的表示式，主要用到生成函数的方法，主要难点是幂级数的计算。

求解

Catalan数的递推关系满足：
$$c_n=\sum _{j=0}^{n-1}{c}_j{c}_{n-1-j},(n\ge 1,c_0=c_1=1)$$
记
$$C(x)=\sum _{n\ge 0}{c}_n{x}^n,$$
于是有：
$$\begin{array}{rl}\sum _{n\ge 1}{c}_n{x}^n& =C(x)-1\\ & =\sum _{n\ge 1}\sum _{j=0}^{n-1}{c}_j{c}_{n-1-j}{x}^n\\ & =x\sum _{n\ge 1}\sum _{j=0}^{n-1}{c}_j{c}_{n-1-j}{x}^{n-1}\\ & =x\sum _{n\ge 0}\sum _{j=0}^n{c}_j{c}_{n-j}{x}^n\\ & =x\sum _{j\ge 0}{c}_j{x}^j\sum _{n\ge j}{c}_{n-j}{x}^{n-j}\\ & =xC(x)\cdot C(x)\end{array}$$
即：
$$x{C}^2(x)-C(x)+1=0,$$
立即解得：
$$C(x)=\frac{1}{2x}(1\pm \sqrt{1-4x}),$$
由幂级数收敛条件可知
$$C(x)=\frac{1}{2x}(1-\sqrt{1-4x}),$$
展开上式：
$$\begin{array}{rl}C(x)& =\frac{1}{2x}\left[1-\sum _{n\ge 0}\left(\genfrac{}{}{0ex}{}{\frac{1}{2}}{n}\right)(-4{)}^n{x}^n\right]\\ & =\frac{1}{2x}\left[1-1-\sum _{n\ge 1}\left(\genfrac{}{}{0ex}{}{\frac{1}{2}}{n}\right)(-4{)}^n{x}^n\right]\\ & =\frac{1}{2x}\left[-\sum _{n\ge 1}\left(\genfrac{}{}{0ex}{}{\frac{1}{2}}{n}\right)(-4{)}^n{x}^n\right]\\ & =-\frac{1}{2}\sum _{n\ge 1}\left(\genfrac{}{}{0ex}{}{\frac{1}{2}}{n}\right)(-4{)}^n{x}^{n-1}\\ & =-\frac{1}{2}\sum _{n\ge 0}\left(\genfrac{}{}{0ex}{}{\frac{1}{2}}{n+1}\right)(-4{)}^{n+1}{x}^n\end{array}$$
于是有：
$$c_n=-\frac{1}{2}\left(\genfrac{}{}{0ex}{}{\frac{1}{2}}{n+1}\right)(-4{)}^{n+1}$$
下面利用牛顿二项式定理化简上面的结果：
$$\begin{array}{rl}{c}_n& =-\frac{1}{2}\left(\genfrac{}{}{0ex}{}{\frac{1}{2}}{n+1}\right)(-4{)}^{n+1}\\ & =(-1{)}^{n+2}{2}^{2n+1}\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdots \left(\frac{1}{2}-n-1+1\right)}{(n+1)!}\\ & =(-1{)}^{n+2-n}{2}^{2n+1-n-1}\frac{(2n-1)!!}{(n+1)!}\\ & =\frac{2^n(2n-1)!!(2n)!!}{(n+1)!(n!2^n)}\\ & =\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\end{array}$$