喜闻乐见的公式编辑测试环节，联系太多了，所以肯定写不完
联系下降幂，上升幂，幂
[(x)^n=x(x+1)...(x+n-1) \
(x)_n=x(x-1)...(x-n+1)
]
[sumlimits_{k=1}^{n}S_1(n,k)x^k=(x)^n \
sumlimits_{k=1}^{n}(-1)^{n-k}S_1(n,k)x^k=(x)_n \
sumlimits_{k=1}^{n}S_2(n,k)(x)_k=(x)_n=x^n
]
递推关系对偶
[S_1(n,k)=(n-1)cdot S_1(n-1,k)+S_1(n-1,k-1) \
S_1(0,0)=1 \
S_1(n,0)=S_1(0,n)=0 for ngeq 1
]
[S_2(n,k)=kcdot S_2(n-1,k)+S_2(n-1,k-1)\
S_2(0,0)=1 \
S_2(n,0)=S_2(0,n)=0 for ngeq 1
]
生成函数对偶
[sumlimits_{n=0}^{infty}S_2(n,k)frac{x^n}{n!}=frac{(e^x-1)^k}{k!}
]
[sumlimits_{n=0}^{infty}S_1(n,k)frac{x^n}{n!}=frac{(ln(x+1))^k}{k!}
]
矩阵
let
[
A=(a_{ij})_{n imes n}=[ (-1)^{i-j}S_1(i,j) ]_{n imes n}\
B=(b_{ij})_{n imes n}=(S_2(i,j))_{n imes n}
]
then
[AB=BA=I
]
注：这里需要对(i>n)的斯特林数做定义，具体的定义方式我这里找不到了
还有一些
让(A(x),B(x))分别为({a_n}_{n=0}^{infty})和({b_n}_{n=0}^{infty})的指数生成函数，以下三命题等价
[forall ngeq 0 , b_n=sumlimits_{i=0}^{infty}S_2(n,i)a_i\
forall ngeq 0 , a_n=sumlimits_{i=0}^{infty}(-1)^{n-i}S_1(n,i)b_i\
B(x)=A(e^x-1) quad ext{i.e.} quad A(x)=B( ln(1+x) )
]
[
]