[seqfan] Re: connected edge-rooted graphs
Max Alekseyev
maxale at gmail.com
Thu May 3 22:15:50 CEST 2018
Richard, your formulae make sense.
I've just extended A126133 (and soon will upload a b-file with 40 terms),
and so you can get as many terms for the sequence B.
Regards,
Max
On Thu, May 3, 2018 at 5:16 AM, Richard J. Mathar <mathar at mpia-hd.mpg.de>
wrote:
> If A(x) is the the generating function of connected edge-rooted graphs
> with n nodes,
> and A126122(x) the generating function of edge-rooted graphs with n nodes
> (not necessarily connected) (is it?),
> and A000088(x) the generating function of graphs with n nodes
> (not necessarily connected), is then
> A(x) * A000088(x) = A126122(x) ?
>
> A -> 0, 1, 2, 10, 56, 477, 5879, 117729, 4014125, 242887444,
> 26562628943... (n>=1)
>
> Can we combine in the same fashion by dividing
> A000664 (number of graphs with n edges)
> A126133 (number of edge-rooted unlabeled graphs with n edges)
> to get B (number of connected edge-rooted unlabeled graphs with n edges) ?
> B -> 1, 1, 4, 10, 32, 101, 346, 1220,... (n>=1)
> n=1: the connected graph with 1 edge
> n=2: the connected linear graph with 2 edges
> n=3: the triangular graph (1 choice),
> the star graph (1 choice)
> the linear graph with 3 edges (2 choices)
> n=4: the quadrangle (1 choice)
> the star tree graph (1 choice)
> the linear tree graph with 4 edges (2 choices)
> the triangle with a protruding edge (3 choices)
> the tree of 2-methyl-Butane (3 choices)
>
> RJM
>
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