Do we know whether P^#P = NP^#P? I thought the relation between P using a #P-oracle and NP using a #P-oracle is still unknown (or equivalently the relation between P^PP and NP^PP).

Recently, I have read in a journal article that P^#P = NP^#P (citing Toda's publication "PP is as hard as the polynomial time hierarchy"). As far as I remember this relation was posed as an open question in Toda's paper.

I am very confused at the moment and I would be glad if someone could clarify this situation. Thanks a lot.
This is an open question. If it is to be true the consequences would imply the collapse of the ${\bf Counting-Hierarchy}$, ${\bf CH}$ for short. In the paper "On the closure properties of #P in the context of PF o #" by Ogihara, Toda, Watanabe and Thierauf" on proposition 2.1 it is stated that 

${\bf NP ^{PP}} \subseteq {\bf P^{\#P^{[1]}}} $ if and only if ${\bf CH} = {\bf P^{\#P^{[1]}}} $ if and only if ${\bf FP ^{CH}} \subseteq {\bf FP^{\#P^{[1]}}} $.