スターリング数の逆行列

by nomura · 2024年5月12日

スターリング数の逆行列
第1種スターリング数と第2種スターリング数の積について以下が成り立つ。

(1)

\begin{align*} \delta_{nj} & =\sum_{k=0}^{n}S_{1}\left(n,k\right)S_{2}\left(k,j\right)\\ & =\sum_{k=j}^{n}S_{1}\left(n,k\right)S_{2}\left(k,j\right) \end{align*}

(2)

\begin{align*} \delta_{nj} & =\sum_{k=0}^{n}S_{2}\left(n,k\right)S_{1}\left(k,j\right)\\ & =\sum_{k=j}^{n}S_{2}\left(n,k\right)S_{1}\left(k,j\right) \end{align*}

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\(S_{1}\left(n,k\right)\)は第1種スターリング数
\(S_{2}\left(n,k\right)\)は第2種スターリング数

第1種スターリング数と第2種スターリング数は互いに逆行列のような関係にある。

(1)

\begin{align*} \sum_{k=0}^{2}S_{1}\left(2,k\right)S_{2}\left(k,2\right) & =\sum_{k=2}^{2}S_{1}\left(2,k\right)S_{2}\left(k,2\right)\\ & =S_{1}\left(2,2\right)S_{2}\left(2,2\right)\\ & =1\cdot1\\ & =1 \end{align*}

(2)

\[ \sum_{k=0}^{2}S_{1}\left(2,k\right)S_{2}\left(k,3\right)=0 \]

(3)

\begin{align*} \sum_{k=0}^{2}S_{2}\left(2,k\right)S_{1}\left(k,2\right) & =\sum_{k=2}^{2}S_{2}\left(2,k\right)S_{1}\left(k,2\right)\\ & =S_{2}\left(2,2\right)S_{1}\left(2,2\right)\\ & =1\cdot1\\ & =1 \end{align*}

(4)

\[ \sum_{k=0}^{2}S_{2}\left(2,k\right)S_{1}\left(k,3\right)=0 \]

(1)

\begin{align*} P\left(x,n\right) & =\sum_{k=0}^{n}S_{1}\left(n,k\right)x^{k}\\ & =\sum_{k=0}^{n}S_{1}\left(n,k\right)\sum_{j=0}^{k}S_{2}\left(k,j\right)P\left(x,j\right)\\ & =\sum_{j=0}^{n}\sum_{k=j}^{n}S_{1}\left(n,k\right)S_{2}\left(k,j\right)P\left(x,j\right)\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n}S_{1}\left(n,k\right)S_{2}\left(k,j\right)P\left(x,j\right) \end{align*} これより、
\begin{align*} \delta_{nj} & =\sum_{k=0}^{n}S_{1}\left(n,k\right)S_{2}\left(k,j\right)\\ & =\sum_{k=j}^{n}S_{1}\left(n,k\right)S_{2}\left(k,j\right) \end{align*}

(2)

\begin{align*} x^{n} & =\sum_{k=0}^{n}S_{2}\left(n,k\right)P\left(x,k\right)\\ & =\sum_{k=0}^{n}S_{2}\left(n,k\right)\sum_{j=0}^{k}S_{1}\left(k,j\right)x^{j}\\ & =\sum_{k=0}^{n}\sum_{j=0}^{k}S_{2}\left(n,k\right)S_{1}\left(k,j\right)x^{j}\\ & =\sum_{j=0}^{n}\sum_{k=j}^{n}S_{2}\left(n,k\right)S_{1}\left(k,j\right)x^{j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n}S_{2}\left(n,k\right)S_{1}\left(k,j\right)x^{j} \end{align*} これより、
\begin{align*} \delta_{nj} & =\sum_{k=0}^{n}S_{2}\left(n,k\right)S_{1}\left(k,j\right)\\ & =\sum_{k=j}^{n}S_{2}\left(n,k\right)S_{1}\left(k,j\right) \end{align*}