Rubik's cube function I'm thinking in a function and if it's possible to solve that.
I have been playing with the cube using the following move: $R U L' U'.$
I notice that the cube solves itself with a certain number of moves: 28 moves to $2\times2\times2$ and 112 to $3\times3\times3.$ (If the cube already is solved).
Then I'm trying to create a formula to calculate the number of moves for the another cubes like $4\times4\times4, 5\times5\times5, 6\times6\times6\dots$
Since:

$x \rightarrow y$
$1 \rightarrow 0$
$2 \rightarrow 28$
$3 \rightarrow 112$
$4 \rightarrow z$

Where x is the number of the cube $(2 = 2\times2\times2, 3 = 3\times3\times3 \dots)$ and y is the amount of moves, I came up with two formulas: 28*$((x-1)^{(x-1)})$ and 28*$(x-1)^2$. Thus, the value for $z$ could be 252 or 756,
My questions are:


*

*Are any of these formulas correct? If so, which one?

*Can be my reasoning corret about the formulas?

*If I'm wrong, answer me why!


Have you tried it on a $4\times4\times4$ cube?  If not there are online rubiks cubes you can play around with.
But to answer your question my guess is that for any cube size $3$ and up you'll solve it again in $112$ moves.  That is because R U L' U' doesn't touch any of the "inner" edges so that means larger cubes will all behave like a $3\times3\times3$ cube in this case.
