多项式递推trick

数论trick

\[F_n(x)=\prod_{i=0}^n(1-p^ix) \]

可以考虑

\[F_n(px)=\prod_{i=0}^{n}(1-p^{i+1}x)=\prod_{i=1}^{n+1}(1-p^ix)=F_n(x)\frac{1-p^{n+1}x}{1-x} \]

即

\[(1-x)F_n(px)=(1-p^{n+1}x)F_n(x) \]

考虑\([x^k]\)

\[p^k[x^k]F_n-p^{k-1}[x^{k-1}]F_n=[x^k]F_n-p^{n+1}[x^{k-1}]F_n\\ (p^k-1)[x^k]F_n=(p^{k-1}-p^{n+1})[x^{k-1}]F_n\\ [x^k]F_n=\frac{p^{k-1}-p^{n+1}}{p^k-1}[x^{k-1}]F_n \]