Superfactorial and Hyperfactorial
Variants of Factorial
Factorial
Recall that the factorial function $n!$ is the product of the first $n$ natural numbers:
\[n!=\prod_{k=1}^{n}k=1\times2\times3\times\cdots\times n\]While there is much to discuss about the factorial function, this post concerns itself with two particular variations of the factorial. Namely the superfactorial and the hyperfactorial.
Superfactorial
We denote the superfactorial of $n$ as $n{$}$. It is defined1 as the product of the first $n$ factorials:
\[n$=\prod_{k=1}^{n}k!=1!\times2!\times3!\times\cdots\times n!\]Hyperfactorial
The hyperfactorial of $n$ is denoted $H(n)$ and is defined as:
\[H(n)=\prod_{k=1}^{n}k^k=1^1\times2^2\times3^3\times\cdots\times n^n\]Calculator
Below is a calculator for computing these factorials. Note that it accepts non-integer arguments as well, in which case it uses the generalizations discussed in the next section.
Domains & Generalizations
The domain of these three functions is the natural numbers, which includes $0$. As such, it’s important to note that $0!,0{$}$, and $H(0)$ all equal $1$ by their above definitions because they are empty products, i.e. a product of 0 factors.
On another note, you might recall that the factorial function can be generalized to the complex plane via analytic continuation, resulting in the gamma function $\Gamma(z)$.
Like the factorial, the superfactorial and hyperfactorial can also be generalized to the complex numbers, resulting in the Barnes G-function $G(z)$, and K-function $K(z)$ respectively.
We can summarize this information with the following table:
| Notation | Name | Definition | OEIS Entry | Analytic Continuation |
|---|---|---|---|---|
| $n!$ | Factorial | $\prod_{k=1}^{n}k$ | A000142 | $\Gamma(n+1)$ gamma function |
| $n{$}$ | Superfactorial | $\prod_{k=1}^{n}k!$ | A000178 | $G(n+2)$ Barnes G-function |
| $H(n)$ | Hyperfactorial | $\prod_{k=1}^{n}k^k$ | A002109 | $K(n+1)$ K-function |
Factorial Identity
We can relate all three of the discussed factorial variants with the following identity:
\[n{$}\cdot H(n)=n!^{n+1}\]Proof
We can prove the above statement, which we'll call $P(n)$, by induction: $$P(n)\equiv n{$}\cdot H(n)=n!^{n+1}$$ First we multiply both sides of the equation by $(n+1)!(n+1)^{n+1}$: $$\begin{align}n{$}\cdot H(n)&=n!^{n+1}\\ (n+1)!(n+1)^{n+1}& \ \ \ \ \ (n+1)!(n+1)^{n+1} \end{align}$$ Now let's simplify the left hand side first. Notice that $n{\$}\cdot(n+1)!=(n+1){\$}$ and that $H(n)\cdot(n+1)^{n+1}=H(n+1)$. From this the left hand side simply becomes: $$(n+1){$}\cdot H(n+1)$$ Now let's deal with right hand side. Notice that the expression can be rewritten as: $$\begin{align} n!^{n+1} \color{green}{(n+1)!}(n+1)^{n+1}&=n!^{n+1}\color{green}{n!(n+1)}(n+1)^{n+1}\\ &=n!^{n+2}(n+1)^{n+2}\\ &=(n+1)!^{n+2}\\ \end{align}$$ Putting the right and left hand sides back together we can see that we just proved $P(n+1)$: $$P(n+1)\equiv(n+1){$}\cdot H(n+1)=(n+1)!^{n+2}$$ However $P(n+1)$ was proved under the assumption that $P(n)$ was true. Thus: $$P(n)\implies P(n+1)$$ But, notice that $P(0)$ is true: $$\begin{align} P(0)&\equiv 0{$}\cdot H(0)=(0!)^{0+1}&\\ &\equiv1\cdot1=1&\text{(empty product is $1$)}\\ &\equiv T& \end{align}$$ We thus have by induction: $$\begin{align} &P(n)\implies P(n+1)\\ &P(0)\\ \therefore\ &\hline{\forall n\in \mathbb{N},\ P(n)}\\ \end{align}$$This identity also holds more generally for the analytic continuations of these functions, although we’ll refrain from trying to prove this here:
\[G(n+1)K(n)=\Gamma(n)^{n}\]