等幂和的组合数通项

tommywong

这次绝对要逮到它！！！

$\displaystyle s_m=\sum_{i=1}^n i^{m-1},c_m=C_n^m$

$s_1=c_1$

$s_2=c_1^2-c_2=c_1s_1-c_2$

$s_3=c_1^3-2c_1c_2+c_3=c_1s_2-c_2s_1+c_3$

$s_4=c_1^4-3c_1^2c_2+2c_1c_3+c_2^2-c_4=c_1s_3-c_2s_2+c_3s_1-c_4$

tommywong

找到了！！！哈哈哈！！！

$\displaystyle (-1)^m s_m=\sum \frac{(r_1+r_2+...+r_m)!}{r_1!r_2!...r_m!} \prod_{i=1}^m (-c_i)^{r_i}$

$\displaystyle (-1)^m \sum_{i=1}^n i^{m-1}=\sum \frac{(r_1+r_2+...+r_m)!}{r_1!r_2!...r_m!} \prod_{i=1}^m (-C_n^i)^{r_i}$

tommywong

咦...

$\displaystyle n,\frac{1}{2}n^2+\frac{1}{2}n,\frac{1}{6}n^3+\frac{1}{2}n^2+\frac{1}{3}n,\frac{1}{24}n^4+\frac{1}{4}n^3+\frac{11}{24}n^2+\frac{1}{4}n$

我白痴啊！等幂和是这样......

$\displaystyle \frac{1}{2}n^2 +\frac{1}{2} n,\frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n,\frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2$