f(x)=∑i=0naixi
f(x)=∑i=0nf(i)∏j=0,j≤n,j=ii−jx−j
f(x+m)dieiA(x)B(x)[xt]A(x)B(x)[xt+n]A(x)B(x)f(t+m)=i=0∑nf(i)j=0,j≤n,j=i∏i−jx+m−j=i=0∑n(x+m−n−1)!(x+m−i)i!(−1)n−i(n−i)!f(i)(x+m)!=i!(−1)n−i(n−i)!f(i)=m−n+i1=i=0∑ndixi=i=0∑neixi=i=0∑tdiet−i=i=0∑ti!(−1)n−i(n−i)!(m−n+t−i))f(i)=i=0∑tdiet−i=i=0∑ti!(−1)n−i(n−i)!(m+t−i))f(i)=(x+t)!(x+t−n−1)!f(t+m)=(t+m−n−1)!(t+m)![xt+n]A(x)B(x)=[xt+n]A(x)B(x)i=t+m−n∏t+mi
g(x)[xi][g(x)]k[xi][g−1(x)]k(ik)(ik+i−1)=1−x1=(ik+i−1)=(ik)(−1)i=i!(k−i)!k!=(i−1k)ik−i+1=i!(k−1)!(k+i−1)!=(i−1k+i−2)ik+i−1
(ik)=i!(k−i)!k!=i!k(k−1)...(k−i+1)≡i!k′(k′−1)...(k′−i+1)=i!(k′−i)!k′!≡(ik′)
(kn)=(k−1n−1)+(kn−1)g(x)f(x)=−x+1=a0+a1x+a2x2
f(x)=i=0∑n
ln1−xV1=−ln(1−xV)=−i=1∑ixVi=i=1∑ixVi
Bell_Number
Bn+1=i=0∑n(in)Bi
F(x)n!Bn+1ex⋅F(x)exex+cF(x)F(x)=i=0∑i!Bixi=i=0∑n(n−i)!1i!Bi=F′(x)=F(x)F′(x)=ln(F(x))=eex+c=eex−1
1011100101111010100001011000011010111001+1000011010111111100111110−76−1+6/8×2−8−7/210
fi表示i个点联通的方案数,gi表示任意一图
gngngn(n−1)!gnG′(x)F′(x)F(x)=2(2n)=i=1∑n(i−1n−1)fign−i=i=1∑n(i−1)!(n−i)!(n−1)!fign−i=i=1∑n(i−1)!fi(n−i)!gn−i=F′(x)G(x)=G(x)G′(x)=lnG(x)+c
\begin{align}
\sum_{S\subset T}\mu(\prod_{i\in S}i)\varphi(\prod_{i\in S}i)&=0&,k=0\
\sum_{S\subset T}\mu(\prod_{i\in S}i)\varphi(\prod_{i\in S}i)&=\sum_{S\subset T}(-1)^{|S|}\prod_{i\in S}(i-1)&,k=1\
&=\sum_{S\subset T}\prod_{i\in S}(-(i-1))\
f_{p}(x)&=1-(p-1)x^{p}\
ans&=\prod_{p}f_{p}(x)|{x=1}-1\
\sum{S\subset T}\mu(\prod_{i\in S}i)\varphi(\prod_{i\in S}i)&=\sum_{S\subset T}g_{S}\prod_{i\in S}(-(i-1))
&,k=2\
\end{align}