Consider the set of all topologically inequivalent polyhedral graphs with $k$ edges, the number of which is given by Sloan sequence A002840 (1,0,1,2,2,4,12,22,58,158,448)).

Now define a topological form parameter as $\beta:= (\text{number of vertices}=v)/(k+2)$ and consider the distribution of the polyhedral graphs with $k$ edges as a function of $\beta$. Due to duality the distribution is symmetric about $\beta=1/2$. Due to the fact that for a planar graph $k \le 3v-6$, the distribution vanishes outside the interval $\beta \in [1/3, 2/3]$.

Now a natural question is whether this distribution tends to a limiting distribution when the number of edges tends to infinity. Is it known whether such a limiting distribution exists - or will it be singular, i.e. concentrated with smaller and smaller variance around $\beta=1/2$, as numerical data seem to suggest? Is there any nontrivial limit distribution theorem by means of rescaling?

EDIT: Some numerical data can be found under Using these data gives the following values for the standard deviation of the distribution $p(\beta)$ as a function of the number of edges: $\sigma(k=21) = 0.029895922, \sigma(k=27) = 0,027943943, \sigma(k=33) = 0,02625827$.

  • It's not really important for the answer, but I suggest that you make explicit the offset by 6 in the sequence and briefly explain the exact meaning of "polyhedra" and "inequivalent": it takes a bit of work to understand what is being asked. – Victor Protsak May 30 '10 at 1:53

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