I need to know quantity of two types of combinations - reflecting and inverse (concerning the center). Reflecting mean, that if first is $+$, then last is $+$, and if second is $-$, then penultimate is $-$, and so on. Inverse mean, that if first is $+$, then last is $-$, and if second is $-$, then penultimate is $+$, and so on. For example:
2) $+-=-+$ (inverse)
4) $+--+=-++-$ (reflecting), $++--=--++$, $+-+-=-+-+$ (inverse)
But $+-+-$ is special case of 2, so we exclude it. For better understanding:
$++=++++=++++++=\cdots$
$+-=+-+-=+-+-+-=\cdots$
$++--=++--++--=++--++--++--=\cdots$
6) $++--++=--++--$, $+-++-+=-+--+-$, $+----+=-++++-$ (reflecting)
$+++---=---+++$, $++-+--=--+-++$, $+-+-+-=-+-+-+$, $+--++-=-++--+$ (inverse)
Here $+-+-+-$ is also special case of 2.
So if $q(m)$ - quantity of combinations (reflecting and inverse), not including special cases, $$q(2)=1, q(4)=2, q(6)=6$$ But that doesn't mean, that $q(m)=(m/2)!$
Also with special cases their quantity equal to $$\frac{2^n-2}{2}$$ for reflecting and $$\frac{2^n}{2}$$ for inverse, so summary quantity equal to $2^n-1$.
How can I calculate their quantity (without special cases)?