We consider a matrix
$$A:=\begin{pmatrix} 0 & b & 0 &f \\a & 0 & e & 0 \\ 0 & d & 0 & h \\ c& 0 & g & 0 \end{pmatrix}.$$
This matrix has the interesting property that if you multiply it once with iself, it will have entries only where at the moment it has zero entries, in fact
$$A^2= \left( \begin{matrix} a b+c f & 0 & b e+f g & 0 \\ 0 & a b+d e & 0 & a f+e h \\ a d+c h & 0 & d e+g h & 0 \\ 0 & b c+d g & 0 & c f+g h \\ \end{matrix} \right)$$
and by multiplying with $A$ again
$$A^3=\left( \begin{array}{cccc} 0 & b (a b+c f)+d (b e+f g) & 0 & f (a b+c f)+h (b e+f g) \\ a (a b+d e)+c (a f+e h) & 0 & e (a b+d e)+g (a f+e h) & 0 \\ 0 & b (a d+c h)+d (d e+g h) & 0 & f (a d+c h)+h (d e+g h) \\ a (b c+d g)+c (c f+g h) & 0 & e (b c+d g)+g (c f+g h) & 0 \\ \end{array} \right)$$
the entries are back to where we started from.
I am wondering now, if there is a way to write an explicit (so not recursive) formula for the entries $A_{21}$ and $A_{41}$ of $A^{2n+1}$ and $n \in \mathbb N$?
