[seqfan] Re: Partition into Stroke of the star graph
Max Alekseyev
maxale at gmail.com
Thu May 4 17:55:17 CEST 2023
As I understand from Christian messages, there is no concern about
correctness of A089243, except for the case n=1.
I agree that value for n=1 can be disputable, but this really boils down to
what we understand under the star graph in this case. It is clearly the
graph composed just of a single edge connecting two vertices. If one of
them is labeled as the star center, then we have the current
value A089243(1)=2. However, if the center is not distinguished, then the
value would be 1.
Perhaps, it's worth adding a comment to the sequence explaining this issue.
Regards,
Max
On Thu, May 4, 2023 at 11:13 AM Neil Sloane <njasloane at gmail.com> wrote:
> Concerning A089243, I will ask Max Alekseyev to reply.
>
> Best regards
> Neil
>
> Neil J. A. Sloane, Chairman, OEIS Foundation.
> Also Visiting Scientist, Math. Dept., Rutgers University,
> Email: njasloane at gmail.com
>
>
>
> On Wed, May 3, 2023 at 11:40 PM Christian Sievers <seqfan at duvers.de>
> wrote:
>
> > Some remarks and results:
> >
> > 1) The value of A089243(1)=2 can be disputed.
> >
> > We can only distinguish the two stroke sets for a star graph with one
> > edge if the center node is distinguished. In all other cases (even n=0)
> > it is obvious which point is the center.
> >
> > 2) In the mail, what does this mean?
> >
> > > 1, 2, 3, 4, 9, 22, 61, 200, Off SET 0 0 means Phi
> > ^^^^^^^^^^^
> > It's not from the OEIS entry.
> >
> > 3) Also from the mail:
> >
> > > The definition is for the case of directed , labelled , mirror
> > > symmetric exist I computed all cases that exist eight
> >
> > We can certainly consider variations of the series, like undirected
> > strokes (then we act on sets of sets of size 2, instead of sets of
> > pairs, and count each 2-stroke possibility only once)
> > and/or only consider stroke sets equivalent if one can be obtained from
> > the other by a rotation and not allow reflections (so replace the
> > dihedral group by a cyclic group),
> > but I don't see what labeled vs unlabeled should mean.
> > Without labels I can't differentiate the nodes, and then the symmetries
> > become meaningless.
> >
> > I computed the cases that make sense to me, here are the results
> > (code at the end of the mail):
> >
> > # as before
> > gap> List([1..10],i->f(i,true,true));
> > [ 2, 3, 4, 9, 22, 61, 200, 689, 3054, 12110 ]
> >
> > # only rotations are equivalent
> > gap> List([1..10],i->f(i,true,false));
> > [ 2, 3, 6, 12, 34, 98, 374, 1290, 5962, 23768 ]
> >
> > # undirected
> > gap> List([1..10],i->f(i,false,true));
> > [ 1, 2, 2, 5, 6, 17, 27, 83, 185, 608 ]
> >
> > # undirected and only rotations
> > gap> List([1..10],i->f(i,false,false));
> > [ 1, 2, 2, 5, 6, 18, 34, 108, 294, 984 ]
> >
> > One might argue that in the undirected case the maximal stroke condition
> > allows at most one 1-stroke. Then the results (code not given here) are
> > [ 1, 1, 1, 2, 3, 5, 11, 17, 65, 79 ]
> > with reflection and rotations and
> > [ 1, 1, 1, 2, 3, 5, 15, 18, 105, 105 ]
> > with only rotations considered equivalent.
> >
> > Of these, only the original series is in the OEIS.
> >
> >
> > Best
> > Christian
> > --
> >
> > f := function(n,dir,refl)
> > local s, g, mult, act, h, res;
> >
> > if dir then
> > mult := 2;
> > act := OnSetsTuples;
> > else
> > mult := 1;
> > act := OnSetsSets;
> > fi;
> >
> > s := SymmetricGroup( n );
> >
> > if refl then
> > g := DihedralGroup( IsPermGroup, 2*n);
> > else
> > g := CyclicGroup( IsPermGroup, n);
> > fi;
> >
> > h := function( l )
> > local tss,x;
> > tss := Set([1..l], i->[2*i-1, 2*i]); # a set of l two-strokes
> > x := Orbit( s, tss, act ); # all such sets
> > return Size( OrbitsDomain( g, x, act ) );
> > end;
> >
> > res := mult * Sum( [0..QuoInt(n-1,2)], h );
> > if IsEvenInt(n) then
> > res := res + h( n/2 );
> > fi;
> > return res;
> > end;
> >
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
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