[seqfan] Re: When are stereoisomers counted in A000628?
origami.experience at gmail.com
Sat Feb 11 21:22:15 CET 2017
I think we all agree that the comment should be rephrased. I've came up
with something a little better but I still think there is a better way
describe the enumeration in question:
Steric (three-dimensional) trees can be represented with regular
(two-dimensional) trees provided that stereoisomers are then taken into
account: this means acting the group S_4 on the stereocenters (nodes with 4
different substituents) of each tree and counting all of the resulting
On Thu, Feb 9, 2017 at 2:42 PM, <bradklee at gmail.com> wrote:
> Yes, but your word "symmetric" still confuses me. When describing symmetry
> of a geometric model, I would say invariant / covariant with regard to a
> particular representation, usually irreducible.
> With A000628, I am also slightly confused by the comment, should it rather
> say: "acts invariantly at each node" or "means that the invariant
> transformation group of each node is A4 rather than the full S4"?
> A picture or ASCII drawing might help to clarify the meaning.
> > On Feb 9, 2017, at 5:55 AM, Richard J. Mathar <mathar at mpia-hd.mpg.de>
> > With reference to http://list.seqfan.eu/pipermail/seqfan/2017-
> > The phrase is not useful in both cases, because it is not important with
> > group you act on the molecule (you can act with any geometric group on
> > molecule if you wish) but whether that molecule is symmetric (an
> > or non-symmetric upon action of the symmetry group.
> > What the comment means to say is (?) that in A000628 molecules are
> counted if
> > they are not symmetric under actions of S_4, whereas they are counted in
> > A000602 if they are not symmetric under actions of A_4.
> > --
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