Superfactorial

The ${\displaystyle n}$th superfactorial ${\displaystyle {\mathit {sf}}(n)}$ may be defined as:[1]

$${\displaystyle {\begin{aligned}{\mathit {sf}}(n)&=1!\cdot 2!\cdot \cdots n!=\prod _{i=1}^{n}i!=n!\cdot {\mathit {sf}}(n-1)\\&=1^{n}\cdot 2^{n-1}\cdot \cdots n=\prod _{i=1}^{n}i^{n+1-i}.\\\end{aligned}}}$$

Following the usual convention for the empty product, the superfactorial of 0 is 1. The sequence of superfactorials, beginning with ${\displaystyle {\mathit {sf}}(0)=1}$, is:[1]

Just as the factorials can be continuously interpolated by the gamma function, the superfactorials can be continuously interpolated by the Barnes G-function.[2]

According to an analogue of Wilson's theorem on the behavior of factorials modulo prime numbers, when ${\displaystyle p}$ is an odd prime number

$${\displaystyle {\mathit {sf}}(p-1)\equiv (p-1)!!{\pmod {p}},}$$

where ${\displaystyle$ !!} is the notation for the double factorial.[3]

For every integer ${\displaystyle k}$, the number ${\displaystyle {\mathit {sf}}(4k)/(2k)!}$ is a square number. This may be expressed as stating that, in the formula for ${\displaystyle {\mathit {sf}}(4k)}$ as a product of factorials, omitting one of the factorials (the middle one, ${\displaystyle (2k)!}$) results in a square product.[4] Additionally, if any ${\displaystyle n+1}$ integers are given, the product of their pairwise differences is always a multiple of ${\displaystyle {\mathit {sf}}(n)}$, and equals the superfactorial when the given numbers are consecutive.[1]

• Weisstein, Eric W., "Superfactorial", MathWorld