Assume we have $n$ crosses ( $\times$ ) and $m$ zeroes ( 0 ). I want to find in how many ways we can arrange them up to cyclic permutations for any $n,m$. For instance, for $n = 3$ and $m = 3$ I believe we have four cases,
$0\; 0\;0 \times\times\,\times$
$0\; 0 \times 0 \times\times$
$0\; 0 \times \times\, 0\, \times$
$0 \times 0\times 0 \,\times$
If $n = 2$ (or $m = 2$), then I think we have $\lceil \frac{n + m - 1}{2} \rceil$ possibilities. My best guess was to obtain a general formula by recursion and then hope to be able to sum whatever comes out, but haven't managed so far. Any ideas?