A two-sided $n$-omino is a $1$-by-$n$ board of $n$ squares with each square ($2n$ in all because of the two sides) colored with one of $p$ given colors (squares on opposite sides may be colored differently). How many nonequivalent two-sided $n$-ominoes are there?
I don't know how to do this exercise because I think that it is not possible for $n$ because when specifying the table for the cyclical index I don't know if a given offset will visit all $n$ elements or not.
Why do I think that it is impossible?
Let say that $n=5$ and we consider rotation by $2$ elements. At the begining, let point first element
$$\underline{1},2,3,4,5 $$
Then we rotate by 2 elements $$1,2,\underline{3},4,5 $$ And again $$1,2,3,4,\underline{5} $$ $$1,\underline{2},3,4,5 $$ $$1,2,3,\underline{4},5 $$ $$\underline{1},2,3,4,5 $$
So we have $1$ cycle with size $5$. If I consider just $n$ I don't know which and how many cycles this rotation could create...
Can you help me?