Superfactorials

[Aras14HD]

Description

Superfactorials now have been requested multiple times. They are defined:

$$n\$ = \prod_{i=1}^n i! = G(n+2)$$

A simple notation, that is common enough to warrant its own command. The G here is an analytical continuation, so unlike multitermials and subfactorials (both of which I cannot find one for) it can relatively easily support decimals. One consideration to take is efficiency, we will need to approximate way earlier.

There is a sterling-like approximation:

$$\ln G(1+z) ∼ z^2\left(\frac{1}{2}\ln z-\frac{3}{4}\right)+\frac{1}{2}\ln(2\pi)z-\frac{1}{12}\ln z+\zeta'(-1)+\sum_{k=1}^n\frac{B_{2k+2}}{4k(k+1)z^{2k}}+O\left(\frac{1}{z^{2n+2}}\right)$$

Where B_k is the k-th Bernulli number approximated by:

$$B_{2n} ∼ (-1)^{n-1}4\sqrt{\pi n}\left(\frac{n}{\pi e}\right)^{2n}$$

If that proves too complex (and inefficient) there is a simpler form:

$$\ln G(1+z) ∼ z^2\left(\frac{1}{2}\ln z-\frac{3}{4}\right)+\frac{1}{2}\ln(2\pi)z-\frac{1}{12}\ln z+\zeta'(-1)+O\left(\frac{1}{z}\right)$$

Reasoning

This has been requested here:

• https://www.reddit.com/r/unexpectedfactorial/comments/1lo5e38/comment/n0kar1q
• https://www.reddit.com/r/unexpectedfactorial/comments/1ln30oj/petition_to_add_superfactorial/ (whole post)

Resources

• https://en.wikipedia.org/wiki/Barnes_G-function
• https://mathworld.wolfram.com/BarnesG-Function.html
• https://mathworld.wolfram.com/BernoulliNumber.html
• https://mathworld.wolfram.com/Superfactorial.html

[tolik518]

I haven't found anything online after a quick google, but I can imagine multi-superfactorials might also be a topic - like with multi-termials.

[Aras14HD]

It is a little more ambiguous what multisuperfactorial means:

$$\begin{aligned} 5\$\$ &= 5! \cdot 3! \cdot 1! \\ &\text{or maybe inner} \\ 5\$\$ &= 5!! \cdot 4!! \cdot 3!! \cdot 2!! \cdot 1!! \\ &\text{or combined} \\ 5\$\$ &= 5!! \cdot 3!! \cdot 1!! \end{aligned}$$

Also, there will most definitely be no analytical continuation anywhere. Even an approximation I would have to figure out myself, so it would be limited to only smaller integers.

[tolik518]

You're right. While for me it would make sense it to be

$$\begin{aligned} 5\$\$ &= 5! \cdot 3! \cdot 1! \\ \end{aligned}$$

it might mean something different for others.

Also, there will most definitely be no analytical continuation anywhere. Even an approximation I would have to figure out myself, so it would be limited to only smaller integers.

Yeah I also don't expect too much information on that matter (or rather completely no information).

Edit: Oopps closed the ticket by accident