Is $coNP^{\#P}=NP^{\#P}=P^{\#P}$? By http://www.cs.umd.edu/~jkatz/complexity/relativization.pdf

If $A$ is a PSPACE-complete language, $P^{A}=NP^{A}$.

If $B$ is a deterministic polynomial-time oracle, $P^{B}\ne NP^{B}$ (assuming $P\ne NP$).

$PP$ is the class of decision problems analog for $\#P$ and $P\subseteq PP\subseteq PSPACE$, 

but neither $P=PP$ nor $PP=PSAPCE$ is known. But is it true that

$coNP^{\#P}=NP^{\#P}=P^{\#P}$?
It is an open problem in complexity theory for many years if $\mathsf{PH}^{\mathsf{\#P}}$ collapse,
where $\mathsf{PH}$ is the polynomial time hierarchy. It is also an open problem to construct
an oracle to separate $\mathsf{P}^{\mathsf{\#P}}$ from $\mathsf{PSPACE}$.