The Details:
The astrolabacus is a puzzle said to have $$A_3:=\frac{18!\times 2^{18}}{3!^6}=35,972,733,468,672,000$$ possible positions. Here is a GIF of the puzzle moving:
There are six segments that each contain six coloured balls: three per coloured half segment. Those segments can be moved as above and there are loops of $18$ balls that one can slide into different positions. There are four colours, like so:
A solution to the puzzle is described here.
Definition 1: The $n$-astrolabacus is an astrolabacus where, instead of $3$ balls per half segment, there are $n$ balls.
Let $A_n$ be the number of different positions of the $n$-astrolabacus.
The Question:
What is $A_n$ in terms of $n\in\Bbb N$?
This sequence is not in the OEIS.
My Attempt:
Well, when $n=3$, quoting this article, we have that
[t]he puzzle is symmetrical, and remains so even when it changes its shape ($180$ degree rotational symmetry around a central axis). The balls therefore occur in pairs which remain opposite each other. There are four colours (red, blue, green, yellow), and the tunnels are coloured in such a way that every pair occurs exactly once.
There are $18$ pairs coloured balls (sic) which can be permuted in $18!$ ways. The balls of a pair can also be swapped, so there are (at most) $18!\times 2^{18}$ ways to arrange them all. There are however indistinguishable ball pairs - there are $6$ pairs of colours, all occurring three times.
This gives the number quoted in the details.
Building on this, I conjecture that
$$A_n=\frac{2^{6n}(6n)!}{n!^6}.$$
Is this correct? It seems like it's just a matter of replacing $18$ by $6n$ in the argument above.
The expression $\frac{2^{6n}(6n)!}{n!^6}$ is an integer since $n!$ divides any product of $n$ consecutive integers.
Please help :)