[seqfan] Re: Another planetary sequence
Rick Shepherd
rlshepherd2 at gmail.com
Sun Sep 11 22:44:14 CEST 2016
That probably effectively describes one such simplification. (Disclaimers
before this goes far: I'm neither a combinatorialist nor graph theorist --
although I audited *part* of a graduate combinatorics course once (not the
related graph theory course)).
I would like as general as possible (too): Really n bodies; e.g., not
necessarily just one "star". One body could be orbiting multiple bodies. A
second body could be orbiting multiple other bodies. Maybe both groups are
orbiting another group of "stars". Also, would like not to restrict moons
of moons of... other than by n.
(not really necessarily caring (yet) which are gravitationally stable or
not)
There seem to me to be several possible sequences here.
Thanks for your reply.
Rick
On Sep 11, 2016 4:20 PM, "Olivier Gerard" <olivier.gerard at gmail.com> wrote:
> On Sun, Sep 11, 2016 at 9:12 PM, Rick Shepherd <rlshepherd2 at gmail.com>
> wrote:
>
> >
> > A category of planet-related sequences of interest (in general to me, and
> > combinatorially) is number of types of planetary/star/moon orbital
> systems,
> > probably under various simplifying assumptions (the 3-body problem having
> > not been solved yet).
> >
> > [...]
> >
>
>
> > For simplicity, maybe start with counting all orbital situations that can
> > occur with all orbits elliptical and coplanar... Moons can have moons,
> etc.
> > Include rotations or not. Retrograde orbits or not. Well-defining
> > equivalence classes of orbital systems may be a bit tricky still.
> >
> > (I'm deliberately ignoring Figure-8 orbits and such for now.)
> >
> >
> What you describe looks to me essentially like rooted planar trees of
> height at most 2
> (if we discard moons of moons):
>
> The star as the root, the planets as nodes of depth 1, the moons as leaves,
> ...
>
> The order of the planet is fixed by the planarity and a reading order
> convention.
>
> Including retrograde orbits is just like adding binary coloring to the
> edges,
> including rotation is binary coloring of the nodes while chosing
> arbitrarily
> one of the colours for the root.
>
> Am I missing something ?
>
> Regards,
>
> Olivier
>
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>
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