Are all sets of $n$, s.t. $R(m)^n=I$, where $R(m)$ is any sequence of $m$ moves on a Rubik's cube and $I$ is the identity operator, known?

Question

I've written a program that finds the number of times, $n$, one must apply any operation $R_i(m)$, which consists of $m$ single moves/turns/elementary operations on a Rubik's cube, s.t. $R_i(m)^n=I$, the identity operator. I've chosen to include moving the middle sections as well, so that the legal elementary operations are all 18 possible combinations of $"\!\text{rotate \{1}^{\text{st}}\vee \text{2}^{\text{nd}}\vee \text{3}^{\text{rd}} \} \;\;\{\text{row} \vee \text{column} \vee \text{page} \} \;\frac{\pi}{2}\;\{\text{clockwise} \vee \text{counter-clockwise}\}.\!"$

For instance, if we start from a solved cube and, using Singmaster notation, apply the operation $R(2)=LU'$, we must do this $n = 63$ times before we come back to the exact same solved cube (exact meaning that the cube isn't allowed to be rotated relative to the initial one).

In the first figure below, we see the $n$s for $10^4$ operations consisting of two ($m=2$) randomly chosen moves. Considering that there are $18^m$ different possible operations consisting of $m$ moves, very few values of $n$ are allowed; they, along with the values for $m=1,3$, are \begin{align*} a_1&=\{4\} \\ a_2&=\{1,2,4,8,12,63,105,126,210,1260\}\\ a_3&=\{4,8,12,20,24,30,36,60,72,80,84,90,120,132,168,180,190,210,228,240,252,330,360,390,396,420,560,630,720,840,1190,1260,1560,9240\} \end{align*} (OEIS doens't find anything), and the frequencies of them (as well as for $m=3,4$) are shown in the second figure. I know that $a_2$ is complete (there are no other possible $n$ for $m=2$), as I brute forced all possibilities (which is doable for a computer for $m=1,2,3$, but not much more).

Q: Have all such lists $a_m$ (as well as their relative frequencies) been worked out?

Here are some bonus questions, which needn't be addressed for an answer to be accepted, but which I nonetheless find interesting:

To make the third figure, I averaged all $n$ for a given $m$.

BQ1: Why are $n$ for odd $m$ generally smaller than for even $m$? This somehow seems counter-intuitive.

BQ2: If one splits up the data for $m$ even and odd into two separate graphs, they seem to approximately follow $\sqrt{m}$. Why? (I guess this could be calculated directly if the answer to Q is affirmative.)

Answer 1

Regarding the main question: I suppose "not exactly". However, since every cube can be solved in 20 moves or less, one would "only" have to do the computations up to $m=20$. For $m=20$ the question simply degenerates to

• What are the possible orders of elements of the Rubik group?
• How many elements of a given order exist?

I am sure these two questions have been answered somewhere, and in fact it is not too difficult to compute them manually, mainly using that the group is in the well-known manner of index $2\cdot 3$ in $(S_{12}\rtimes C_2^{12})\times (S_{8}\rtimes C_3^8)$.