We know that counting the number of solutions to $3-SAT$ is the canonical $\#P-Complete$ problem. Equivalently, it is the canonical $PH-Hard$ problem.
However, counting the number of solutions to $3-SAT$ that have weight at most $\log_2n$ (that is, have at most $\log_2n$ variables assigned to True) is a problem in $P$.
If we want to look at the complexity of counting solutions to $3-SAT$ with the maximum weight of an assignment, say $w$, as a parameter, it seems that:
$w = \mathcal{O}(\log_2 n):$ the complexity is $P$
$w = \mathcal{O}(n):$ Complexity is $PH-Hard$
It seems fair to tell that:
$w = \mathcal{O}(\log_2^kn):$ the complexity of counting solutions to $3-SAT$ is complete (or maybe just hard) for $\Sigma^P_{k-1}$.
where $\Sigma^P_{k-1}$ is the $k-1$ level of the polynomial hierarchy.
Can anyone please help me see why the above should be true, or otherwise ? Any reference would also greatly help.