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I guess it would be called #P-Space but I have found only one article vaguely mentioning it. How about the counting version of EXP-TIME-Complete, NEXP-Complete as well as EXP-SPACE-Complete problems? Is there any previous work that can one cite in regards to this or any type of inclusion or exclusion like Toda's Theorem?

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You are asking a lot in one question! – Tsuyoshi Ito Aug 27 '11 at 0:50
#PSPACE is the same as the class of functions which can be computed in polynomial space (FPSPACE). – Tsuyoshi Ito Aug 27 '11 at 0:58
@Tsuyoshi This is true. However, most of the questions asked if not all, can be rephrased as a single general question: Are there counting classes for classes higher than $NP$ (as one can note in the definition of #$P$) and do known results apply? – chazisop Aug 27 '11 at 20:59
@Tayfun Pay: I'm not entirely sure what you mean for deterministic classes like PSPACE, EXP, EXPSPACE. The notion of "number of solutions" is usually closely tied to nondeterminism--since then you can ask about the number of accepting paths--or existential quantifiers/projections. In the case of PSPACE of course you can use the alternating quantifiers definition--but then you have to specify which quantifiers you want to count over--or the fact that NPSPACE=PSPACE. – Joshua Grochow Aug 28 '11 at 19:10
As several comments mentioned, it is not totally clear what you would want to mean for #PSPACE. The best bet would be to take the padded-up analog of #L which is well studied. As #L is contained in DSPACE(log^2 n), this would imply that #PSPACE=PSPACE, as @TsuyoshiIto mentioned above. (I am ignoring here the immaterial formal distinction between decision problems and functions.) – Noam Nov 26 '12 at 2:14

The number of satisfying assignments to a boolean formula equals the number of valid quantifications of the formula. The inductive proof is quite elegant. So #P = #PSpace.

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Isn't this covered by Tsuyoshi and Noam's comments above? – Huck Bennett Aug 19 '14 at 23:20
Is this what you really mean? If #P = #PSPACE, doesn't this imply that PSPACE $\subseteq$ P$^{\#\mathrm{P}}$ ? I don't believe this is known. – Peter Shor Aug 20 '14 at 1:13
@PeterShor I am fairly certain Daniel means this But my (unverified) guess is that a #PSPACE-complete problem is to count the number of satisfying assignments of a fixed QBF, not count the number of satisfiable quantifications of a given CNF. – Sasho Nikolov Aug 20 '14 at 2:30

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