2項係数の1項間漸化式
2項係数は以下の1項間漸化式を満たす。
(1)
\[ C(x+1,y)=\frac{x+1}{x+1-y}C(x,y) \](2)
\[ C(x-1,y)=\frac{x-y}{x}C(x,y) \](3)
\[ C(x,y+1)=\frac{x-y}{y+1}C(x,y) \](4)
\[ C(x,y-1)=\frac{y}{x-y+1}C(x,y) \](5)
\[ C(x+1,y+1)=\frac{x+1}{y+1}C(x,y) \](6)
\[ C(x+1,y-1)=\frac{(x+1)y}{(x-y+1)(x-y+2)}C(x,y) \](7)
\[ C(x-1,y+1)=\frac{(x-y-1)(x-y)}{x(y+1)}C(x,y) \](8)
\[ C(x-1,y-1)=\frac{y}{x}C(x,y) \](1)
\begin{align*} C(x+1,y) & =\frac{(x+1)!}{y!(x+1-y)!}\\ & =\frac{x+1}{x+1-y}\frac{x!}{y!(x-y)!}\\ & =\frac{x+1}{x+1-y}C(x,y) \end{align*}(2)
\begin{align*} C(x-1,y) & =\frac{(x-1)!}{y!(x-1-y)!}\\ & =\frac{x-y}{x}\frac{x!}{y!(x-y)!}\\ & =\frac{x-y}{x}C(x,y) \end{align*}(3)
\begin{align*} C(x,y+1) & =\frac{x!}{(y+1)!(x-y-1)!}\\ & =\frac{x-y}{y+1}\frac{x!}{y!(x-y)!}\\ & =\frac{x-y}{y+1}C(x,y) \end{align*}(4)
\begin{align*} C(x,y-1) & =\frac{x!}{(y-1)!(x-y+1)!}\\ & =\frac{y}{x-y+1}\frac{x!}{y!(x-y)!}\\ & =\frac{y}{x-y+1}C(x,y) \end{align*}(5)
\begin{align*} C(x+1,y+1) & =\frac{(x+1)!}{(y+1)!(x-y)!}\\ & =\frac{x+1}{y+1}\frac{x!}{y!(x-y)!}\\ & =\frac{x+1}{y+1}C(x,y) \end{align*}(6)
\begin{align*} C(x+1,y-1) & =\frac{(x+1)!}{(y-1)!(x-y+2)!}\\ & =\frac{(x+1)y}{(x-y+1)(x-y+2)}\frac{x!}{y!(x-y)!}\\ & =\frac{(x+1)y}{(x-y+1)(x-y+2)}C(x,y) \end{align*}(7)
\begin{align*} C(x-1,y+1) & =\frac{(x-1)!}{(y+1)!(x-y-2)!}\\ & =\frac{(x-y-1)(x-y)}{x(y+1)}\frac{x!}{y!(x-y)!}\\ & =\frac{(x-y-1)(x-y)}{x(y+1)}C(x,y) \end{align*}(8)
\begin{align*} C(x-1,y-1) & =\frac{(x-1)!}{(y-1)!(x-y)!}\\ & =\frac{y}{x}\frac{x!}{y!(x-y)!}\\ & =\frac{y}{x}C(x,y) \end{align*}
ページ情報
| タイトル | 2項係数の1項間漸化式 |
| URL | https://www.nomuramath.com/p8gtr0e4/ |
| SNSボタン |
ディクソンの等式
\[
\sum_{k=-a}^{a}(-1)^{k}C(a+b,a+k)C(b+c,b+k)C(c+a,c+k)=\frac{(a+b+c)!}{a!b!c!}
\]
ファンデルモンドの畳み込み定理と第1引数の畳み込み
\[
\sum_{j=0}^{k}C(x,j)C(y,k-j)=C(x+y,k)
\]
2項係数の半分までの総和
\[
\sum_{k=0}^{n-1}C\left(2n-1,k\right)=2^{2n-2}
\]
2項変換と交代2項変換の逆変換
\[
a_{n}=\sum_{k=0}^{n}(-1)^{n-k}C(n,k)b_{k}
\]