How to derive number of polyhedra with $n$ vertices?

Question

I was thinking about Euler's Formula recently, after noticing a particularly nice proof of it through reverse-induction on graphs, and when thinking about it, it occurred to me that counting the number of polyhedra with $n$ vertices doesn't seem to me to be at all simple...

I found this sequence on the OEIS which counts just this, but I was wondering if anyone could give an explanation as to whether there is a simple function of $n$ which gives us this and, if so, how to derive such a function. Similarly, if that's not possible, how would one go about counting polyhedra in a systematic manner?

For example, I can, in a sense, only count the number of polyhedra with 4 vertices because of my own prior knowledge and Euler's Formula, but I would have no idea where to begin with 63 vertices... If the explanation or reasoning is highly technical, I would appreciate (having a relatively small understanding of geometry) a simplified derivation at least.

Answer 1

Having looked this up, on and off today, I have found out the following facts, which essentially resolve my question and I thought others might be interested in knowing (I certainly didn't!):

• It was noted some time ago that all polyhedra can be projected onto the plane (e.g. stereographic projection) and that they are 3-connected planar graphs. A more recent and non-trivial observation was that, conversely, every 3-connected planar graph can be realised as a convex polyhedron!
• Such graphs are called polyhedral graphs and the number of them, given a number of vertices, can be checked by computer systematically, which is how we have the sequence mentioned above.
• However, there is no known formula that can enumerate the number of (non-isomorphic) such polyhedral graphs for a given $n=|V|$ (at least in well-known literature and survey papers).
• The smallest polyhedron which is not Hamiltonian was discovered using these computer searches and is called the Herschel Graph.

Hope this helps!

EDIT: There's also the existence of Tutte embeddings for just these graphs, which is highly pleasing.