# 遞進階乘與遞降階乘

（重定向自下降阶乘幂

## 定义

### 遞進階乘

${\displaystyle x^{\overline {n}}=x(x+1)(x+2)\cdots (x+n-1)={\frac {(x+n-1)!}{(x-1)!}}}$

### 遞降階乘

${\displaystyle x^{\underline {n}}=x(x-1)(x-2)\cdots (x-n+1)={\frac {x!}{(x-n)!}}}$

### 两者的关系

${\displaystyle x^{\overline {n}}=(x+n-1)^{\underline {n}}}$

${\displaystyle 1^{\overline {n}}=n^{\underline {n}}=n!}$

## 擴展

### 零次幂

${\displaystyle x^{\overline {0}}=x^{\underline {0}}=1}$

### 实数

${\displaystyle x^{\overline {n}}={\frac {\Gamma (x+n)}{\Gamma (x)}}\quad x,x+n\neq 0,-1,-2,\cdots }$

${\displaystyle x^{\underline {n}}={\frac {\Gamma (x+1)}{\Gamma (x-n+1)}}\quad x,x+n\neq -1,-2,-3,\cdots }$

## 特性

${\displaystyle {\frac {x^{\overline {n}}}{n!}}={x+n-1 \choose n}}$
${\displaystyle {\frac {x^{\underline {n}}}{n!}}={x \choose n}}$

${\displaystyle n|x^{\overline {n}}}$
${\displaystyle n|x^{\underline {n}}}$

n=4 ，遞進階乘与遞降階乘必定能表达为一个完全平方数减1，即

${\displaystyle x^{\overline {4}}=k^{2}-1}$
${\displaystyle x^{\underline {4}}=k^{2}-1}$

${\displaystyle (a+b)^{\overline {n}}=\sum _{r=0}^{n}{n \choose r}a^{\overline {n-r}}b^{\overline {r}}}$
${\displaystyle (a+b)^{\underline {n}}=\sum _{r=0}^{n}{n \choose r}a^{\underline {n-r}}b^{\underline {r}}}$

${\displaystyle x^{\underline {m}}x^{\underline {n}}=\sum _{k=0}^{m}{m \choose k}{n \choose k}k!\,x^{\underline {m+n-k}}}$

## 一般化

${\displaystyle [f(x)]^{k/h}=f(x)\cdot f(x+h)\cdot f(x+2h)\cdots f(x+(k-1)h)}$
${\displaystyle [f(x)]^{k/-h}=f(x)\cdot f(x-h)\cdot f(x-2h)\cdots f(x-(k-1)h)}$

## 与亚微积分的關係

${\displaystyle \Delta x^{\underline {k}}=kx^{\underline {k-1}}\,}$

${\displaystyle {\frac {\partial }{\partial x}}{x^{k}}=kx^{k-1}\,}$

## 参考文献

• Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren E., Concrete Mathematics: A Foundation for Computer Science, 1988, ISBN 0-201-14236-8.
• Olver, Peter J., Classical Invariant Theory, Cambridge University Press, 1999, ISBN 0521558212.