When does "X is NP-complete" imply "#X is #P-complete"? Let $X$ denote a (decision) problem in NP and let #$X$ denote its counting version.

Under what conditions is it known that "X is NP-complete" $\implies$ "#X is #P-complete"?

Of course the existence of a parsimonious reduction is one such condition, but this is obvious and the only such condition of which I am aware.  The ultimate goal would be to show that no condition is needed.

Formally speaking, one should start with the counting problem #$X$ defined by a function $f : \{0,1\}^* \to \mathbb{N}$ and then define the decision problem $X$ on an input string $s$ as $f(s) \ne 0$?
The most recent paper on this question seems to be:

Noam Livne, A note on #P-completeness of NP-witnessing relations, 
Information Processing Letters, Volume 109, Issue 5, 15 February 2009, Pages 259–261
http://www.sciencedirect.com/science/article/pii/S0020019008003141

which gives some sufficient conditions.

Interestingly the introduction states "To date, all known NP complete sets have a defining relation which is #P complete", so the answer to Suresh's comment is "no examples are known".