$1-x$で割る事は累積和を取ることと対応
\begin{align}
\frac{1}{1-x} &= 1 + x + x^2 + x^3 + \cdots = \sum_{i = 0}^{\infty} x^i \\
\frac{1}{(1-x)^2} &= 1 + 2x + 3x^2 + 4x^3 + \cdots = \sum_{i = 0}^{\infty} (i+1) x^i \\
\frac{1}{(1-x)^3} &= 1 + 3x + 6x^2 + 10x^3 + \cdots = \sum_{i = 0}^{\infty} \frac{(i+1)(i+2)}{2} x^i \\
\frac{1}{(1-x)^k} &= \binom{k}{0} + \binom{k}{1} x + \binom{k+1}{2} x^2 + \cdots = \sum_{i = 0}^{\infty} \binom{k+i-1}{i} x^i
\end{align}
二項係数の定義
\[
(1+x)^k = \sum_{i = 0}^{k} \binom{k}{i} x^i
\]
分割数と五角数定理
\[
\prod_{ k = 1 }^\infty \frac{1}{1-x^k} = \frac{1}{\sum_{n = -\infty}^{\infty} (-1)^n x^{n(3n-1)/2}}
\]
$\displaystyle \frac{f(x)}{g(x)} \lbrack x^k \rbrack$
は$N = \deg(f)+\deg(g)$とすれば$O(N \log N \log k)$で求まる