？ - 洛谷仓库

省流：

f_i=\sum\limits_{j}f_jf_{i-j-1}\\ g_i=\sum\limits_{j}2f_jg_{i-j-1}

首先有：

\begin{aligned} F(x)&=\sum\limits_{i\ge 0}f_ix^i\\ &=1+\sum\limits_{i\ge 1}f_ix^i\\ &=1+\sum\limits_{i\ge 1}\sum\limits_{0\le j<i}f_jx^jf_{i-j-1}x^{i-j-1}x\\ &=1+x\sum\limits_{i\ge 0}\sum\limits_{0\le j\le i}f_jx^jf_{i-j}x^{i-j}\qquad\qquad i\gets i-1，提出\ x\\ &=1+x\sum\limits_{i\ge 0}\sum\limits_{0\le j\le i}f_jx^if_{i-j}\\ &=1+x\sum\limits_{i\ge 0}x^i\sum\limits_{0\le j\le i}f_jf_{i-j}\\ &=1+x[F(x)]^2 \end{aligned}

此时我们有：

\begin{aligned} F(x)&=1+x[F(x)]^2\\ xF(x)^2-F(x)+1&=0\\ F(x)&=\dfrac{1\pm \sqrt{1-4x}}{2x} \end{aligned}

取

F(x)=\dfrac{1-\sqrt{1-4x}}{2x}

同样的，可以得到

G(x)

：

\begin{aligned} G(x)&=\sum\limits_{i\ge 0}g_ix^i\\ &=x+\sum\limits_{i\ge 2} \sum\limits_{0\le j<i}2f_jx^jg_{i-j-1}x^{i-j-1}x\\ &=x+2x\sum\limits_{i\ge 0}\sum\limits_{0\le j\le i}f_jx^jg_{i-j}x^{i-j}\\ &=x+2x\sum\limits_{i\ge 0}\sum\limits_{0\le j\le i}f_jx^ig_{i-j}\\ &=x+2x\sum\limits_{i\ge 0}x^i\sum\limits_{0\le j\le i}f_jg_{i-j}\\ &=x+2xF(x)G(x) \end{aligned}

有：

\begin{aligned} G(x)&=x+2xF(x)G(x)\\ G(x)-2xF(x)G(x)&=x\\ (1-2xF(x))G(x)&=x\\ G(x)&=\dfrac{x}{1-2xF(x)}\\ \end{aligned}

代入

F(x)

有：

\begin{aligned} G(x)&=\dfrac{x}{1-2x\dfrac{1-\sqrt{1-4x}}{2x}}\\ &=\dfrac{x}{1-(1-\sqrt{1-4x})}\\ &=\dfrac{x}{\sqrt{1-4x}}\\ &=x(1-4x)^{-\frac{1}{2}} \end{aligned}

考虑展开. 有：

\begin{aligned} \dfrac{-1}{2}^{\underline n}&=\dfrac{-1}{2}\times\dfrac{-3}{2}\times\cdots\\ &=\prod\limits_{i=0}^{n-1}\dfrac{-1-2i}{2}\\ &=\prod\limits_{i=0}^{n-1}(-1)\left(\dfrac{2i+1}{2}\right)\\ &=(-1)^n\prod\limits_{i=0}^{n-1}\left(\dfrac{2i+1}{2}\right)\\ &=(-1)^n\dfrac{1}{2^n}\prod\limits_{i=0}^{n-1}(2i+1)\\ &=(-1)^n\dfrac{1}{2^n}\prod\limits_{i=1}^{n}(2i-1)\\ &=(-1)^n\dfrac{1}{2^n}(2n-1)!!\\ &=(-1)^n\dfrac{1}{2^n}\dfrac{(2n)!}{(2n)!!}\\ &=(-1)^n\dfrac{1}{2^n}\dfrac{(2n)!}{2^nn!}\\ \end{aligned}

根据牛顿二项式定理，代入，有：

\begin{aligned} G(x)&=x\sum\limits_{i\ge 0}\left(\dfrac{-1}{2}\right)^{\underline i}\dfrac{1}{i!}(-4x)^i\\ &=x\sum\limits_{i\ge 0}(-1)^i(-1)^i\dfrac{1}{2^i}\dfrac{(2i)!}{2^ii!}\dfrac{1}{i!}4^ix^i\\ &=x\sum\limits_{i\ge 0}\dfrac{(2i)!}{(i!)^2}x^i\\ &=\sum\limits_{i\ge 0}\dfrac{(2i)!}{(i!)^2}x^{i+1}\\ &=\sum\limits_{i\ge 1}\binom{2i}{i}x^i \end{aligned}

于是得出

G(x)

关于

F(x)

，差别不大，不想推了.

？