Superfactorials

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:

Resources

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.

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.

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

Gonna write down the derivation digitally later, for now here are the formulas (except float, as I would use the Bernoulli part for them, and digits, as it makes more sense to go directly to tower):

$$\begin{aligned} &\text{Tower} \\ 10^{10^{n}}\$ &\approx 10^{10^{10^{n}}} \\ \left(^{n}10\right)^{a}\$ &\approx \left(^{n+1}10\right)^{a} \\ &\text{Approx} \\ (n-1)\$ &\approx 10^{n^2\frac{l}{2}\ln{10}\div\ln{10}}10^{n^2\frac{1}{2}\ln{k}\div\ln{10}}10^{-n^2\frac{3}{4}\div\ln{10}} \\ & \space \space \space \space 10^{\frac{1}{2}\ln(2\pi)n\div\ln{10}}10^{-\frac{l}{12}\ln{10}\div\ln{10}}10^{-\frac{1}{12}\ln{k}\div\ln{10}}10^{\zeta'(-1)\div\ln{10}} \\ n &= k \cdot 10^l | l \in \mathbb{Z} \end{aligned}$$

For approx, I would for each of the exponents separate whole number from decimals and make the base (the a in a*10^b) with those decimals (10^decimals). Then multiply and adjust going along. Sorta:

exponents.into_iter().map(|x| {let b = x.clone().floor(); (ten.pow(x-b),b)}).fold((1,0), |a,e| adjust((a.0*e.0,a.1+e.1)))

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Superfactorials #211

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tolik518 commented on Jun 30, 2025

Aras14HD commented on Jun 30, 2025

tolik518 commented on Jun 30, 2025

Aras14HD commented on Feb 27, 2026

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