Earlier today I calculated the number of distinct states for the Petaminx (see this post) and since I was already more than halfway to a general formula for any N-minx I figured I'd go the rest of the way.

I present, the general formula for any N-Minx:

nminx(n) = (((20! / 2) * (3^20 / 3)) * (((30! / 2) * (2^30 / 2)))^(n % 2) * (60! / 2)^(((n + (1 - (n % 2))) - 3) / 2) * (60! / (5!^12))^((((n + (1 - (n % 2))) - 3) / 2) * ((((n + (1 - (n % 2))) - 3) / 2) + (n % 2)))) / (60^(1 - (n % 2)))

It is defined as the number of layers to an edge. That is, the Flowerminx / Kilominx is N = 2, the Megaminx is N = 3, the Petaminx is N = 9, etc. It is only valid for N >= 2. The % operator is a modulus so 3 % 2 = 1 and 4 % 2 = 0.

Here are a few interesting values.

Kilominx: nminx(2) = 23563902142421896679424000
Megaminx: nminx(3) = 100669616553523347122516032313645505168688116411019768627200000000000
Master Kilominx: nminx(4) = 9.149 * 10^163
Gigaminx: nminx(5) = 3.647 * 10^263
Teraminx: nminx(7) = 1.151 * 10^573
Petaminx: nminx(9) = 3.164 * 10^996
Examinx: nminx(11) = 7.576 * 10^1533
[...]
Oskarminx: nminx(17) = 4.538 * 10^3829
[...]
Centaminx: nminx(100) = 4.787 * 10^140809

WOW! Great job! I'm blown away!

This is an awesome piece of maths! Thanks for sharing.

Could I ask for a derivation of this formula?

Is it generalisable to find permutations of an n-order version of an m-sided shape? Obviously it would have to assume that it is face turning!

is it further generalisable to vertex or edge turning n-order, m-sided shapes?

I'm sure there's a PhD in there somehow!!

Well it looks complicated but broken down, it's deceptively simple.

The basic idea is to look at how many of each piece type there are as you make the puzzle bigger. The corners stay constant. The middle edge layer goes away for even puzzles, and the edge wings grow linearly with N and the face pieces grow with the square of N.

To make the middle edges go away I just raise that whole term to the 0-power to make it 1 by doing a ^(n % 2).

The exponents make use of % 2 to change the formula slightly for even versus odd edge-lengths.

A bit of algebraic manipulation / simplification could happen in the exponents which would make the whole thing a bit easier to understand.

No, the formula doesn't work for any other puzzle types or geometries. The general algorithm I used to derive the formula works for most other twisty puzzles and geometries but for a lot of puzzles you have to worry about orbits and parity and twist relationships between piece types.

Well I do think there is enough math and theory and practice related to twisty puzzles to support at least one PhD thesis but this formula or the work behind it doesn't even come close.