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(Ans=frac{sumlimits_{i=0}^ni^k(m-1)^{n-i}inom ni}{m^k})
(F(x)=sumlimits_{tge0}frac{x^t}{t!}sumlimits_{i=0}^ni^tinom ni(m-1)^{n-i})
(=sumlimits_{i=0}^ninom ni(m-1)^{n-i}e^{ix})
(=(e^x+m-1)^n)
(现在求[x^k]F(x),我们试用EI介绍的方法进行处理:)
(看成复合形态G(e^x),G(x)=(x+m-1)^n)
(注意G(e^k)=G((e^k-1)+1))
(考虑H(x)=G(x+1)=(x+m)^n)
(nH(x)=(x+m)H'(x))
(nH_i=iH_i+m(i+1)H_{i+1})
(m(i+1)H_{i+1}+(i-n)H_i=0)
(而Q(x)=H(x)\%x^{n+1})
(那么m(i+1)Q_{i+1}+(i-n)Q_i=[i==k](k-n)Q_k)
(Q'(x)(x+m)-nQ(x)=(k-n)m^{n-k}inom nkx^k)
(令P(x)=Q(x-1))
(P'(x)(x+m-1)-nP(x)=(k-n)m^{n-k}inom nk(x-1)^k)
((n-i)P_i=(m-1)(i+1)P_{i+1}+(-1)^{k-i}(n-k)inom nkinom kim^{n-k})
(解出P之后就容易做了)
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