[seqfan] Re: A250001

[Jon Wild]

On Sat, 8 Aug 2015, Benoît Jubin wrote:

> Following the article in QuantaMagazine which raises awareness of 
> A250001, I submitted a sequence (table) which counts the configurations 
> with given numbers of intersections. It is A261070.

I couldn't see your sequence, it doesn't appear to be live yet. But I now 
have reason to believe the 4th term of A250001 is incorrect (it was me who 
hand-computed the terms originally). Instead of a(4) = 168, I believe a(4) 
= 173. I have a program now, which I'm almost confident enough in to 
update and extend the results--I have a few last checks to make--and it 
gets me to n=6.

Perhaps there is a graph theorist who could say whether or not it is easy 
to enumerate the number of embedded edge-coloured graphs using n colours 
such that:

--every vertex is of degree 4
--the edges that meet in each vertex alternate between 2 colours
--the set of edges of any given colour forms a cycle within the graph

The number of vertices in such a graph will be even, and range from 2(n-1) 
to n(n-1).

This will give the number of connected configurations of n circles, which 
is the first step my algorithm uses, before adding any disconnected 
circles or configurations of circles in the regions of the original 
embedded graph. I make this sequence of connected configurations 1, 6, 
112, 15528... (you are allowed to turn the configurations over, i.e. 
right-handed and left-handed embeddings are not counted separately).

The principal difficulty is in determining that each resulting embedded 
graph is in fact drawable using circles! For n=5 (15528 connected 
configurations) I think it very likely that the solutions my program found 
all work. At some point though, as n gets larger this will no longer be 
the case.

Jon Wild