关于(dfrac{1}{(1-z)^c})的幂级数
首先，(G(z)dfrac{1}{1-z})表示(egin{aligned}{sum_{i=0}^ng_i}end{aligned})的生成函数
(c=0,G_0(z)=1z^0+0z^1+0z^2+0z^3+cdots)
(c=1,G_1(z)=1z^0+1z^1+1z^2+1z^3+cdots)
(c=2,G_2(z)=1z^0+2z^1+3z^2+4z^3+cdots)
(c=3,G_3(z)=1z^0+3z^1+6z^2+10z^3+cdots)
(cdotscdotscdotscdotscdotscdotscdots)
容易发现，也很好证明([z^n]G_c(z)=[z^{n-1}]G_c(z)+[z^n]G_{c-1}(z))
(g(c,n)=g(c-1,n)+g(c,n-1))=从((1,0))到((c,n))的只能向右或向上走的方案数(=dbinom{c+n-1}{n})
(egin{aligned}dfrac{1}{(1-z)^c}=sum_{ngeq0}dbinom{c+n-1}{n}end{aligned})
一道题目
给定(n,m,k)，定义(egin{aligned}zeta({a_i})=(m-sum a_i)prod((a_i+k)^2-k^2)end{aligned})
求(egin{aligned}sumzeta({a_i})，forall a_i>0，sum a_i leq mend{aligned})
(egin{aligned}G(z)&=sum_{igeq0}((i+k)^2-k^2)z^i\&=sum_{igeq0}i^2z^i+2ksum_{igeq0}iz^i\&=dfrac{(z+1)z}{(1-z)^3}+dfrac{2kz}{(1-z)^2}end{aligned})
(F(z)=egin{aligned}sum_{igeq0}iz^i=dfrac{z}{(1-z)^2}end{aligned})
(egin{aligned}ANS&=sum_{i=0^*}^m(m-i)[z^i]G(z)^n\&=[z^m]F(z)G(z)^n\&=[z^m]dfrac{z}{(1-z)^2}cdot dfrac{z^n(1+2k+(1-2k)z)^n}{(1-z)^{3n}}\&=[z^{m-n-1}](sum_{igeq0}dbinom{3n+i+1}{i}z^i)(sum_{igeq0}dbinom{n}{i}(1-2k)^i(1+2k)^{n-i}z^i)end{aligned})
*:因为(G(z)^n)没有次数小于(n)的项，所以从(0)开始和从(n)开始一样。