missing sequence? number of embeddings of planar graphs

[Brendan McKay]

You are quite right.  The situation is rather more complicated.

Brendan.

* franktaw at netscape.net <franktaw at netscape.net> [070311 02:16]:
> I don't believe that #1 is any simple transformation of #2.  Consider 
> the graph with a triangle and two additional points.  There are two 
> embeddings of this graph, depending on whether the additional points 
> are on the same side of the triangle.  The general question is quite 
> complex - you have to consider whether the various holes in various 
> components are equivalent or not.
> 
> Franklin T. Adams-Watters
> 
> 
> -----Original Message-----
> From: bdm at cs.anu.edu.au
> 
> I have #2 up to 14 vertices (3807081193879), computed by myself
> and Gunnar Brinkmann.  Presumably #1 follows by the appropriate
> transformation.
> 
> ...
> Brendan.
> 
> * N. J. A. Sloane <njas at research.att.com> [070310 14:25]:
> >
> ...
> >1.  Number of planar graphs on n unlabeled nodes,
> >where we count each graph according to the number
> >of its embeddings into the sphere.
> >
> ...
> >
> >2. Same question for connected planar graphs
> >
> >Could someone work out the first few terms?
> 
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