I'm working in a projective Hilbert space and I have $n$ unit vectors as a basis. I'm trying to find how many unique linear combinations can be made sticking only to adding, or subtracting other basis vectors once.
My naive attempt was simple counting: for our first vector we have $2*n$ options: $+V_i$ or $-V_i$. So we have as we go through: $2n*2(n-1)*...*2 = 2^n*n!$ but this over counts certainly, as $V_a+V_b$ and $-V_a-V_b$ represent the same vector in a projective space. What factor do these duplicates contribute?
Example for clarity: $$n=1: V_0$$ $$n=2: V_0, V_1, V_0+V_1, -V_0+V_1$$ $$n=3: V_0, V_1, V_2, V_0+V_1, V_1+V_2, V_0+V_2,-V_0+V_1, -V_1+V_2, -V_0+V_2, V_0+V_1+V_2, -V_0+V_1+V_2,V_0-V_1+V_2,V_0+V_1-V_2$$ Something like $$-V_0+V_1-V_2$ is simply the negation of the second last term, so it doesnt count.
The coefficient of every basis vector is one of $-1,0,1$, so there are $3^n$ such choices in total. Excluding $0$, there are $3^n-1$ choices. Projectivising means we only double-counted (since we started with a basis), giving the answer $\frac12(3^n-1)$.
Your attempt only counted those with every coefficient nonzero, and counted each combination $n!$ times.