多重階乗同士の関係
多重階乗同士の関係
\(q\in\mathbb{N}_{0}\)とする。
\(q\in\mathbb{N}_{0}\)とする。
(1)
\[ \left(qn+1\right)!_{n}=\left(qn+1\right)!^{n} \](2)
\[ \left(qn+r\right)!^{n}=r!^{n}\frac{\left(qn+r\right)!_{n}}{r!_{n}} \]*
\(x!_{n}\)は多重階乗、\(x!^{n}\)は拡張多重階乗。(1)
\begin{align*} \left(qn+1\right)!_{n} & =n^{q}\frac{\left(q+\frac{1}{n}\right)!}{\left(\frac{1}{n}\right)!}\\ & =n^{\frac{qn+1-1}{n}}\frac{\left(\frac{qn+1}{n}\right)!}{\left(\frac{1}{n}\right)!}\\ & =\left(qn+1\right)!^{n} \end{align*}(2)
\begin{align*} \left(qn+r\right)!^{n} & =r!^{n}\prod_{k=1}^{q}\frac{\left(kn+r\right)!^{n}}{\left(kn+r-n\right)!^{n}}\\ & =r!^{n}\prod_{k=1}^{q}\left(kn+r\right)\\ & =r!^{n}\prod_{k=1}^{q}\frac{\left(kn+r\right)!_{n}}{\left(kn+r-n\right)!_{n}}\\ & =r!^{n}\frac{\left(qn+r\right)!_{n}}{r!_{n}} \end{align*}ページ情報
タイトル | 多重階乗同士の関係 |
URL | https://www.nomuramath.com/hcnmgy5n/ |
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(拡張)多重階乗の逆数和
\[
\sum_{k=0}^{n}\frac{1}{\left(ak+b\right)!_{a}}=\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\left(\frac{\Gamma\left(n+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(n+\frac{b}{a}+1\right)}-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right)
\]
拡張多重階乗の簡単な値
\[
0!^{n}=\frac{1}{\sqrt[n]{n}\left(\frac{1}{n}\right)!}
\]
多重階乗と拡張多重階乗の定義
\[
\left(x\right)!^{n}=n^{\frac{x-1}{n}}\frac{\left(\frac{x}{n}\right)!}{\left(\frac{1}{n}\right)!}
\]
負の多重階乗
\[
\left(-\left(qn+r\right)\right)!_{n}=\frac{\left(-1\right)^{q}}{\left(qn-\left(n-r\right)\right)!_{n}}
\]