【题解】翻转 flip GF 做法

\[\begin{aligned} \frac{F_{n,k}}{k!}&=\frac{1}{k}\sum_{i+j+k=n-1,u+v+w=k-1}\frac{F_{i,u}}{u!}\frac{F_{j,v}}{v!}\frac{F_{k,w}}{w!}\\ \frac{F_{n,k}}{k!}&=\frac{1}{k}\sum_{i+j+k=n-1} [y^k]yF_iF_jF_k\\ \frac{F_{n,k}}{k!}&=\frac{1}{k}[x^ny^k]xyF^3\\ f_{n,k}\times k&=[x^ny^k]xyF^3\\ y\frac{d F}{d y}&=xyF^3\\ \frac{d F}{F^3}&=x\times dy\\ \int \frac{d F}{F^3}&=\int x\times dy\\ -\frac{1}{2F^2}&=xy+C(x)\\ C(x)&=-\frac{1}{2}+x-\frac{1}{2}x^2\\ F&=\pm(1+x^2-2xy-2x)^{-\frac{1}{2}}\\ F&=\sum_{i=0} x^i(x-2y-2)^{i}\binom{-\frac{1}{2}}{i}\\ F&=\sum_{i=0} x^i\binom{-\frac{1}{2}}{i}\sum_{j}\binom{i}{j}x^j(-2)^{i-j}(y+1)^{i-j}\\ [x^ny^k]F&=\sum_{i=0} \binom{-\frac{1}{2}}{i}\binom{i}{2i-n}(-2)^{2i-n}\binom{2i-n}{k}\\ [x^ny^k]F&=\sum_{i=0} \binom{-\frac{1}{2}}{k}\binom{-\frac{1}{2}-k}{i-k}(-2)^{2i-n}\binom{i-k}{2i-n-k}\\ [x^ny^k]F&=\binom{-\frac{1}{2}}{k}(-2)^{k+k-n}\sum_{j} \binom{-\frac{1}{2}-k}{j}4^{j}[z^{n-k-j}](1+z)^j\\ [x^ny^k]F&=\binom{-\frac{1}{2}}{k}(-2)^{k+k-n}[z^{n-k}]\sum_{j} \binom{-\frac{1}{2}-k}{j}4^{j}(1+z)^jz^{j}\\ [x^ny^k]F&=\binom{-\frac{1}{2}}{k}(-2)^{k+k-n}[z^{n-k}](1+4(1+z)z)^{-\frac{1}{2}-k}\\ [x^ny^k]F&=\binom{-\frac{1}{2}}{k}(-2)^{k+k-n}[z^{n-k}](1+4z+4z^2)^{-\frac{1}{2}-k}\\ [x^ny^k]F&=\binom{-\frac{1}{2}}{k}(-2)^{k+k-n}[z^{n-k}](1+2z)^{-1-2k}\\ [x^ny^k]F&=\binom{-\frac{1}{2}}{k}(-1)^{n}2^{k}\binom{-1-2k}{n-k}\\ [x^ny^k]F&=(2k-1)!!\binom{n+k}{2k} \end{aligned} \]