負の多重階乗

負の多重階乗

\(q\in\mathbb{N}_{0}\)とする。

(1)

\[ \left(-\left(qn+r\right)\right)!_{n}=\frac{(n-r)!_{n}\left(-1\right)^{q}}{\left(n-r\right)\left(qn-\left(n-r\right)\right)!_{n}}\cnd{n\ne r} \]

(2)

\(0<r<n\)のとき

\[ \left(-\left(qn+r\right)\right)!_{n}=\frac{\left(-1\right)^{q}}{\left(qn-\left(n-r\right)\right)!_{n}} \]

(1)

\begin{align*} \left(-\left(qn+r\right)\right)!_{n} & =(n-r)!_{n}\prod_{k=0}^{q}\frac{\left(-\left(nk+r\right)\right)!_{n}}{\left(-\left(nk+r\right)+n\right)!_{n}}\\ & =(n-r)!_{n}\prod_{k=0}^{q}\frac{\left(-\left(nk+r\right)\right)!_{n}}{\left(-\left(nk+r\right)+n\right)\left(-\left(nk+r\right)\right)!_{n}}\\ & =(n-r)!_{n}\prod_{k=0}^{q}\frac{-1}{n\left(k-1\right)+r}\\ & =(n-r)!_{n}\left(-1\right)^{q}\frac{1}{n-r}\prod_{k=1}^{q}\frac{1}{n\left(k-1\right)+r}\\ & =(n-r)!_{n}\left(-1\right)^{q}\frac{1}{n-r}\prod_{k=0}^{q-1}\frac{1}{nk+r}\\ & =\frac{(n-r)!_{n}\left(-1\right)^{q}}{\left(n-r\right)\left(qn-\left(n-r\right)\right)!_{n}} \end{align*}

(2)

\(0<r<n\)のとき\(0<n-r<n\)となるので\(\left(n-r\right)!_{n}=n-r\)となる。

これより、

\begin{align*} \left(-\left(qn+r\right)\right)!_{n} & =(n-r)!_{n}\left(-1\right)^{q}\frac{1}{\left(n-r\right)\left(qn-\left(n-r\right)\right)!_{n}}\cmt{(1)\text{より}}\\ & =\frac{\left(-1\right)^{q}}{\left(qn-\left(n-r\right)\right)!_{n}} \end{align*}

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負の多重階乗

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