a0=1a_0 = 1a0=1 ai=(i+1)∑k=0i−1akai−1−kCi−1ka_i = (i+1)\sum_{k=0}^{i-1}a_ka_{i-1-k}C_{i-1}^kai=(i+1)∑k=0i−1akai−1−kCi−1k
求 ana_nan 通项