[seqfan] Re: A250001

[Benoît Jubin]

Here is the formula I mentionned for T261070(n,1).
Write a(n) = T261070(n,1) and b(n) = T261070(n,0) = A000081(n+1).
All configurations accounting for a(n) can be constructed as follows (and
uniquely so): first draw two intersecting circles, then draw n-2-k circles
surrounding them (0<=k<=n-2). This separates the plane into n+2-k connected
regions, two of which are indistinguishable, where one has to draw the
remaining k non-intersecting circles. Hence the formula:
a(n) = \sum_{k=0}^{n-2} \sum_{n_1+\dots+n_{n+2-k}=k, 0 \leq n_i, n_1 \leq
n_2} \prod_{i=1}^{n+2-k} b(n_i).

On Sat, Aug 8, 2015 at 9:27 PM, Neil Sloane <njasloane at gmail.com> wrote:

> I just approved Benoit's A261070, so that people
> can make updates
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
> On Sat, Aug 8, 2015 at 2:50 PM, Jon Wild <wild at music.mcgill.ca> wrote:
>
> > On Sat, 8 Aug 2015, Benoît Jubin wrote:
> >
> > I submitted a sequence (table) which counts the configurations with given
> >> numbers of intersections. It is A261070.
> >>
> >> I hope T(4,1) is correct, and I think T(n,1) is not to hard to compute
> >> from
> >> T(0,0), T(1,0), ..., T(n-2,0) in a similar way:
> >>
> >
> > By the way Benoît, your T(n,0) should give A00081. For T(4,1) (meaning
> > only one pair of circles out of 4 intersects) I make the answer 15. You
> can
> > see all 15 among the 2nd and 3rd rows of the image
> > https://oeis.org/A241601/a241601.png
> >
> > Jon
> >
> >
> >
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