スターリング数とベルヌーイ数の関係
スターリング数とベルヌーイ数の関係
スターリング数とベルヌーイ数について以下が成り立つ。
第1種スターリング数
第2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
\(B_{k}\)はベルヌーイ数
スターリング数とベルヌーイ数について以下が成り立つ。
第1種スターリング数
(1)
\[ \frac{\left(-1\right)^{m}}{m!}\sum_{k=0}^{m}\left(-1\right)^{k}S_{1}\left(m+1,k+1\right)B_{k}=\frac{1}{m+1} \](2)
\[ \frac{\left(-1\right)^{m-1}}{\left(m-1\right)!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m,j\right)B_{j}=\frac{1}{m+1} \]第2種スターリング数
(3)
\[ B_{m-1}=\sum_{k=1}^{m}\left(-1\right)^{m+k}\frac{\left(k-1\right)!}{k}S_{2}\left(m,k\right) \](4)
\[ B_{m}=\sum_{k=0}^{m}\left(-1\right)^{k}\frac{k!}{k+1}S_{2}\left(m,k\right) \](5)
\[ B_{m}=\sum_{j=1}^{m}\left(-1\right)^{m+j-1}S_{2}\left(m,j\right)\frac{\left(j-1\right)!}{j+1}+\delta_{0,m} \]-
\(S_{1}\left(n,k\right)\)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
\(B_{k}\)はベルヌーイ数
(1)
\(m\in\mathbb{N}_{0}\)とする。\begin{align*} \frac{1}{m+1}\sum_{k=0}^{m+1}\left(-1\right)^{m+k+1}S_{1}\left(m+1,k\right)n^{k} & =\frac{\left(-1\right)^{m+1}P\left(n,m+1\right)}{m+1}\\ & =\frac{Q\left(n,m+1\right)}{m+1}\\ & =\sum_{k=0}^{n-1}Q\left(k+1,m\right)\\ & =\sum_{k=0}^{n-1}\frac{1}{k}Q\left(k,m+1\right)\\ & =\sum_{k=0}^{n-1}\frac{1}{k}\sum_{j=0}^{m+1}\left(-1\right)^{m+j+1}S_{1}\left(m+1,j\right)k^{j}\\ & =\sum_{k=0}^{n-1}\sum_{j=1}^{m+1}\left(-1\right)^{m+j+1}S_{1}\left(m+1,j\right)k^{j-1}\\ & =\sum_{k=0}^{n-1}\sum_{j=0}^{m}\left(-1\right)^{m+j}S_{1}\left(m+1,j+1\right)k^{j}\\ & =\sum_{j=0}^{m}\left(-1\right)^{m+j}S_{1}\left(m+1,j+1\right)\frac{1}{j+1}\left(B_{j+1}\left(n\right)-B_{j+1}\left(0\right)\right)\\ & =\sum_{j=0}^{m}\left(-1\right)^{m+j}S_{1}\left(m+1,j+1\right)\frac{1}{j+1}\left(\sum_{k=0}^{j+1}C\left(j+1,k\right)B_{j+1-k}n^{k}-\sum_{k=0}^{j+1}C\left(j+1,k\right)B_{j+1-k}0^{k}\right)\\ & =\sum_{j=0}^{m}\left(-1\right)^{m+j}S_{1}\left(m+1,j+1\right)\frac{1}{j+1}\left(\sum_{k=0}^{j+1}C\left(j+1,k\right)B_{j+1-k}n^{k}-B_{j+1}\right)\\ & =\sum_{j=0}^{m}\sum_{k=0}^{j+1}\left(-1\right)^{m+j}\frac{1}{j+1}S_{1}\left(m+1,j+1\right)\left(C\left(j+1,k\right)B_{j+1-k}-B_{j+1}\delta_{0k}\right)n^{k} \end{align*} \(n\)の1次の項を比べると、
\begin{align*} \frac{1}{m+1} & =\frac{\left(-1\right)^{m}}{m!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m+1,j+1\right)B_{j} \end{align*}
(2)
\(m\in\mathbb{N}\)とする。\begin{align*} \frac{1}{m+1} & =\frac{\left(-1\right)^{m}}{m!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m+1,j+1\right)B_{j}\\ & =\frac{\left(-1\right)^{m}}{m!}\sum_{j=0}^{m}\left(-1\right)^{j}\left\{ S_{1}\left(m,j\right)-mS_{1}\left(m,j+1\right)\right\} B_{j}\\ & =\frac{\left(-1\right)^{m}}{m!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m,j\right)B_{j}+\frac{\left(-1\right)^{m-1}}{\left(m-1\right)!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m,j+1\right)B_{j}\\ & =\frac{\left(-1\right)^{m}}{m!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m,j\right)B_{j}+\frac{\left(-1\right)^{m-1}}{\left(m-1\right)!}\sum_{j=0}^{m-1}\left(-1\right)^{j}S_{1}\left(m,j+1\right)B_{j}\\ & =\frac{\left(-1\right)^{m}}{m!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m,j\right)B_{j}+\frac{1}{m} \end{align*} これより、
\begin{align*} \frac{\left(-1\right)^{m}}{m!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m,j\right)B_{j} & =\frac{1}{m+1}-\frac{1}{m}\\ & =-\frac{1}{m\left(m+1\right)} \end{align*} となるので、
\[ \frac{\left(-1\right)^{m-1}}{\left(m-1\right)!}\sum_{j=0}^{m}\left(-1\right)^{j}S_{1}\left(m,j\right)B_{j}=\frac{1}{m+1} \] となる。
(3)
\(m\in\mathbb{N}\)とする。\begin{align*} B_{m-1} & =\sum_{k=0}^{m-1}\delta_{m,k+1}B_{k}\\ & =\sum_{k=0}^{m-1}\left(-1\right)^{m+k+1}\delta_{m,k+1}B_{k}\\ & =\sum_{k=0}^{m}\sum_{j=0}^{m}\left(-1\right)^{m+k+1}S_{2}\left(m,j\right)S_{1}\left(j,k+1\right)B_{k}\\ & =\sum_{k=0}^{m}\sum_{j=-1}^{m-1}\left(-1\right)^{m+k+1}S_{2}\left(m,j+1\right)S_{1}\left(j+1,k+1\right)B_{k}\\ & =\sum_{j=-1}^{m-1}\left(-1\right)^{m+1}S_{2}\left(m,j+1\right)\sum_{k=0}^{m}\left(-1\right)^{k}S_{1}\left(j+1,k+1\right)B_{k}\\ & =\sum_{j=-1}^{m-1}\left(-1\right)^{m+1}S_{2}\left(m,j+1\right)\sum_{k=0}^{j}\left(-1\right)^{k}S_{1}\left(j+1,k+1\right)B_{k}\\ & =\sum_{j=0}^{m-1}\left(-1\right)^{m+1}S_{2}\left(m,j+1\right)\sum_{k=0}^{j}\left(-1\right)^{k}S_{1}\left(j+1,k+1\right)B_{k}\\ & =\sum_{j=0}^{m-1}\left(-1\right)^{m+1}S_{2}\left(m,j+1\right)\left(-1\right)^{j}\frac{j!}{j+1}\\ & =\sum_{j=0}^{m-1}\left(-1\right)^{m+j+1}\frac{j!}{j+1}S_{2}\left(m,j+1\right)\\ & =\sum_{j=1}^{m}\left(-1\right)^{m+j}\frac{\left(j-1\right)!}{j}S_{2}\left(m,j\right) \end{align*}
(4)
\begin{align*} B_{m} & =\sum_{k=1}^{m+1}\left(-1\right)^{m+1+k}\frac{\left(k-1\right)!}{k}S_{2}\left(m+1,k\right)\\ & =\sum_{k=1}^{m+1}\left(-1\right)^{m+1+k}\frac{\left(k-1\right)!}{k}\left\{ S_{2}\left(m,k-1\right)+kS_{2}\left(m,k\right)\right\} \\ & =\sum_{k=1}^{m+1}\left(-1\right)^{m+1+k}\frac{\left(k-1\right)!}{k}S_{2}\left(m,k-1\right)+\left(-1\right)^{m+1}\sum_{k=1}^{m+1}\left(-1\right)^{k}\left(k-1\right)!S_{2}\left(m,k\right)\\ & =\sum_{k=1}^{m+1}\left(-1\right)^{m+1+k}\frac{\left(k-1\right)!}{k}S_{2}\left(m,k-1\right)-\left(-1\right)^{m+1}\delta_{1,m}\\ & =\left(-1\right)^{m}\sum_{k=0}^{m}\left(-1\right)^{k}\frac{k!}{k+1}S_{2}\left(m,k\right)-\delta_{1,m}\\ & =\left(-1\right)^{m}\sum_{k=0}^{m}\left(-1\right)^{k}\frac{k!}{k+1}S_{2}\left(m,k\right)-\left(\left(-1\right)^{m}-1\right)B_{m}\\ & =\sum_{k=0}^{m}\left(-1\right)^{k}\frac{k!}{k+1}S_{2}\left(m,k\right) \end{align*}(5)
\(m\in\mathbb{N}_{0}\)とする。\begin{align*} B_{m} & =\sum_{k=0}^{m}\delta_{mk}B_{k}\\ & =\sum_{k=0}^{m}\left(-1\right)^{m+k}\delta_{mk}B_{k}\\ & =\sum_{k=0}^{m}\left(-1\right)^{m+k}\sum_{j=0}^{m}S_{2}\left(m,j\right)S_{1}\left(j,k\right)B_{k}\\ & =\sum_{j=0}^{m}\left(-1\right)^{m}S_{2}\left(m,j\right)\sum_{k=0}^{m}\left(-1\right)^{k}S_{1}\left(j,k\right)B_{k}\\ & =\sum_{j=0}^{m}\left(-1\right)^{m}S_{2}\left(m,j\right)\sum_{k=0}^{j}\left(-1\right)^{k}S_{1}\left(j,k\right)B_{k}\\ & =\sum_{j=1}^{m}\left(-1\right)^{m}S_{2}\left(m,j\right)\sum_{k=0}^{j}\left(-1\right)^{k}S_{1}\left(j,k\right)B_{k}+\delta_{0,m}B_{0}\\ & =\sum_{j=1}^{m}\left(-1\right)^{m}S_{2}\left(m,j\right)\left(-1\right)^{j-1}\frac{\left(j-1\right)!}{j+1}+\delta_{0,m}\\ & =\sum_{j=1}^{m}\left(-1\right)^{m+j-1}S_{2}\left(m,j\right)\frac{\left(j-1\right)!}{j+1}+\delta_{0,m} \end{align*}
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タイトル | スターリング数とベルヌーイ数の関係 |
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スターリング数の母関数
\[
\sum_{n=0}^{\infty}S_{1}\left(n,k\right)\frac{x^{n}}{n!}=\frac{\log^{k}\left(1+x\right)}{k!}
\]
第1種スターリング数の符号
\[
\left|S_{1}\left(n,k\right)\right|=\left(-1\right)^{n+k}S_{1}\left(n,k\right)
\]
スターリング数の解釈
\[
\left(-1\right)^{n+k}S_{1}\left(n,k\right)=\sum_{1\leq a_{1}<a_{2}<\cdots<a_{n-k}\leq n-1}\prod_{j=1}^{n-k}a_{j}
\]
スターリング数の逆行列
\[
\delta_{nj}=\sum_{k=0}^{n}S_{1}\left(n,k\right)S_{2}\left(k,j\right)
\]