So if we take a look at the binary representation of numbers, we can see that the digits follow a pattern:
$$0 = 0000$$ $$1 = 0001$$ $$2 = 0010$$ $$3 = 0011$$ $$4 = 0100$$ $$\dots$$
i.e. the least significant digit goes 0,1,0,1,... and repeats. The 2nd least significant digit goes 0,0,1,1,... and repeats. So I have a formula for the "generating function" of the $i$th least significant digit for the first $2^V$ values:
$$f(i,V) = \left(\prod_{k=0}^{i-1}(1+x^{2^k})\right)x^{2^i}\left(\prod_{k=i+1}^{V}(1+x^{2^k})\right)$$
Can we prove this formula is correct?
UPDATE
Here's an easier version of the formula:
$$f(i,V) = \left(\prod_{k=0}^{V}(1+x^{2^k})\right)\frac{x^{2^i}}{1+x^{2^i}}$$
I would write your generating function as $$\dfrac{x^{2^i}}{1 + x^{2^i}} \prod_{k=0}^V (1 + x^{2^i})$$
Now $P_V(x) = \prod_{k=0}^V (1 + x^{2^i}) = \sum_{j=0}^{2^{V+1}-1} x^j$, since each $j$ is represented in exactly one way as the sum of a set of powers of $2$. On the other hand, $$ \dfrac{x^{2^i}}{1 + x^{2^i}} = x^{2^i} - x^{2 \cdot 2^i} + x^{3 \cdot 2^i} + \ldots$$ In expanding the product of this with $P_V(x)$, $x^m$ occurs once for every way of writing $m = j + r \cdot 2^i$ with $0 \le j < m$, with coefficient $-1$ if $r$ is even and $+1$ if $r$ is odd.
If $j_0$ is the least residue of $m$ mod $2^i$, we can have $r = 1, 2, \ldots, (m - j_0)/2^i$. If $(m-j_0)/2^i$ is odd there is one more odd $r$ than even $r$, so the coefficient of $x^m$ in your generating function is $1$, while if $(m - j_0)/2^i$ is even there are as many odd as even $r$ and the coefficient is $0$. But the $2^i$'s digit of the binary expansion of $m$ is the $1$'s digit in the expansion of $(m - j_0)/2^i$, which is $1$ if and only if $(m - j_0)/2^i$ is odd.