Mathematica has a function called FindInstance that looks for a single solution satisfying a set of specified criteria.

I think the question is whether a constant $c$ exists for which a solution is possible for all choices of $m$, $n$, $a_1,\ldots,a_m$.

I believe $MV(A,K) = \int_{S^1} h_A(\theta) dS_K(\theta)$, where $h_A$ is the support height function of $A$ and $S_K$ is the surface area measure (perimeter measure in this case) of $K$. Clearly if $MV(\cdot,K)=MV(\cdot,K')$ then $S_K=S_{K'}$. The fact that $S_K$ uniquely determines $K$ is a well-studied problem, which if I remember correctly is due to Aleksandrov. I will have to look for the appropriate references, but everything should be in the Handbook of Convex Geometry.

What about a measure with all weight given to $(0,0,0,\ldots)$?

In cases where the optimal packing of circles in a circle yields a centered circles, the two numbers should match. I can't figure out how to identify those N's for which the former happens from the tables at packomania.com, so I can't check if this is true for the putative optima given by packomania.com and by Hopkins et al.

The paper also appears on the arXiv (arxiv.org/abs/1002.0604) and on Torquato's website (cherrypit.princeton.edu/papers/paper-292.pdf).

To be clear, "Barlow packing" means a stacking of close-packed triangular layers in ways other than h.c.p. or f.c.c. It does not, to my knowledge, means packings that are saturated under the strongest notions of saturation.

I believe that's right: you can have pockets of arbitrary (saturated) arrangements as long as the total volume of those pockets have zero density. However, under the most stringent definitions of saturation, you can have only Barlow stackings, and those, I believe have finite local complexity. For ideas about what it means for a packing to be saturated, see Kuperberg "Notions of Denseness" (arXiv:math/9908003).

Also keep in mind that a sphere packing in three dimensions, even if it achieves the highest density and is saturated (in the sense that there is no space to add an extra ball without moving the others) can fail to have finite local complexity. You probably need a stronger definition of saturation.

I'm not familiar in detail with the terms you use (finite local complexity, etc.), but I suspect that not much more can be said for optimal packing of spheres of different radii than can be said for tiling with general tile sets. In the latter case, note that the tiling can end up being as complex as the output from an arbitrary Turing machine (see Wang tiles).

Regarding "in the cases of dimensions 2 and 3 where the optimal packing is known, the sets end up being lattices," I assume you're not talking about a finite number of different radii any more but about a single radius.

@Gerry: You are right. This is a proof that $\tan\gamma=\ldots$ if $\tan(\gamma/2)=\ldots$, rather than the converse. Assuming $\tan(\gamma)=\ldots$, we get merely that $f(\tan(\gamma/2))=f(\tan(\beta/2)\tan(\alpha/2))$, where $f(t)=2t/(1-t^2)$, as you note in your answer.

Oh yeah. The second inequality has more assumptions that I forgot to include $f(0)=g(0)=0$.