奇怪的式子 huangzirui · 2020-09-28 22:58:09 · 个人记录 F(z) = F_0(z)-\dfrac{G(F_0(z))}{G'(F_0(z))}\pmod {z^n} \displaystyle \sum_{n \ge 0} \tbinom{n+m}{n} = \dfrac{1}{(1-x)^{m+1}}