多重階乗の階乗表示
多重階乗の階乗表示
\(q\in\mathbb{N}_{0}\)とする。
\[ \left(qn+r\right)!_{n}=r!_{n}n^{q}\frac{\left(q+\frac{r}{n}\right)!}{\left(\frac{r}{n}\right)!} \]
\(q\in\mathbb{N}_{0}\)とする。
\[ \left(qn+r\right)!_{n}=r!_{n}n^{q}\frac{\left(q+\frac{r}{n}\right)!}{\left(\frac{r}{n}\right)!} \]
*
\(x!_{n}\)は多重階乗。\begin{align*}
\left(qn+r\right)!_{n} & =r!_{n}\prod_{k=1}^{q}\left(kn+r\right)\\
& =r!_{n}n^{q}\prod_{k=1}^{q}\left(k+\frac{r}{n}\right)\\
& =r!_{n}n^{q}\prod_{k=1}^{q}\frac{\left(k+\frac{r}{n}\right)!}{\left(k+\frac{r}{n}-1\right)!}\\
& =r!_{n}n^{q}\frac{\left(q+\frac{r}{n}\right)!}{\left(\frac{r}{n}\right)!}
\end{align*}
ページ情報
| タイトル | 多重階乗の階乗表示 |
| URL | https://www.nomuramath.com/seux06gt/ |
| SNSボタン |
(拡張)多重階乗の逆数和
\[
\sum_{k=0}^{n}\frac{1}{\left(ak+b\right)!_{a}}=\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\left(\frac{\Gamma\left(n+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(n+\frac{b}{a}+1\right)}-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right)
\]
2重階乗の逆数和
\[
\sum_{k=0}^{n}\frac{1}{\left(2k\right)!!}=\sqrt{e}\frac{\Gamma\left(n+1,\frac{1}{2}\right)}{\Gamma\left(n+1\right)}
\]
階乗の多重階乗表示
\[
n!=\prod_{k=0}^{j-1}\left(n-k\right)!_{j}
\]
負の多重階乗
\[
\left(-\left(qn+r\right)\right)!_{n}=\frac{\left(-1\right)^{q}}{\left(qn-\left(n-r\right)\right)!_{n}}
\]