Mathematical expression to represent puzzle box solution sequence I'm trying to add a bit of mathematical flair to a japanese puzzle box solution that I am gifting to my friend for his kids.
4 out of the box's 6 sides have movable parts, 2 of those 4 sides have 2 movement possibilities, for a total of 6 possible moves.
The sequence of these movements is 1:6 (1,2,3,4,5,6), repeated four times, followed by movement 1 in reverse (-1), and finally moves 2:3, for a total of 27 moves.
I thought it might be possible to represent this in sequence notation, or as an algebraic function, but I'm having trouble.
Can anyone suggest a direction for this problem? Thank you.
 A: If these kids are old enough to know powers, i.e. $x^y$, and you want to sneak in some abstract math, the best (arguably standard?) notation might be styled after group theory. 
 First name the moves $A,B,C,D,E,F$ and their inverses (reverses) $A^{-1}, B^{-1},$ etc.  Then your solution is:
$$(ABCDEF)^4 A^{-1} BC$$
Some potential "teachable moments" :) would include -


*

*In normal multiplication (of numbers) the order does not matter, but in this notation the order matters.  I.e. $AB$ means doing $A$ then $B$, and is different from $BA$.  (If they know matrices then they know another example of $AB \neq BA$.)

*Even though these abstract objects are not numbers (they are moves), a "thing" raised to an integral power $N$ (here $N=4$) still means multiplying $N$ of that "thing" in sequence.  Here $(ABCDEF)^4 = (ABCDEF)(ABCDEF)(ABCDEF)(ABCDEF)$.  Note that the order matters (see previous point) and this is not the same as $AAAABBBBCCCCDDDDEEEEFFFF.$

*For numbers $x (\neq 0)$, we write $x^{-1}$ to mean $1/x$ (and you can explain why this notation makes sense: it preserves the rule of $x^m x^n = x^{m+n}$ when e.g. $m=3, n=-1$).  For abstract objects we can still write $A^{-1}$ if multplying $A \times A^{-1}$ (i.e. doing $A$ followed by its inverse) equals do-nothing (equivalent to $1$ being the multiplicative identity).
If they aren't old enough to know powers, convert everything to addition and use instead:
$$4 \times (A+B+C+D+E+F) + (-A) + B+C$$
but again explain that order matters ($A+B \neq B+A$), what $\times$ and $-$ mean when you are dealing with abstract objects (not numbers), etc.
But here's my most sincere recommendation: Do not, under any circumstances, ruin the enjoyment of the puzzle by forcing math on them, making them think: $math \neq fun$!
