[seqfan] Re: [The Tiling List] Re: Coordination sequences for planar nets

Neil Sloane njasloane at gmail.com
Mon Nov 24 15:02:15 CET 2014


Maurizio, Thanks for that correction! Of course you are right.
I can only say that it was very late at night ...

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 Mon, Nov 24, 2014 at 1:38 AM, Maurizio Paolini <paolini at dmf.unicatt.it>
wrote:

> Neil, thanks for citing me in A250120.
>
> However, what I don't get is why you resort to the second difference,
> since already the *first* difference show the periodicity 5:
> 4 (4 6 4 5 5) repeated.
>
> In any case there is a mistake in your "FORMULA" section, since the
> period of the second difference is (2 -2 1 0 -1), its length is of course
> 5, not 4.
>
> Best regards,
> Maurizio
>
> On Mon, Nov 24, 2014 at 12:36:33AM -0500, Neil Sloane wrote:
> > I've fully updated A250120 - thanks to everyone for their contributions.
> >
> > The full list of sequences for the 11 planar nets now looks like this:
> >
> > List of coordination sequences for uniform planar nets: A008458 (the
> planar
> > net 3.3.3.3.3.3), A008486 (6^3), A008574 (4.4.4.4 and also apparently
> > 3.4.6.4), A008576 (4.8.8), A008579 (3.6.3.6), A008706 (3.3.3.4.4),
> A072154
> > (4.6.12), A219529 (3.3.4.3.4), A250120 (3.3.3.3.6), A250122 (3.12.12).
> >
> > Darrah, I just created the last one in your name (mostly to stake your
> > claim to it, also to complete the list). Of course, add more material
> there
> > as it develops.
> > Updates to any of these entries are welcomed.
> >
> > 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 Sun, Nov 23, 2014 at 3:04 PM, Brad Klee <bradklee at gmail.com> wrote:
> >
> > > Hi Neil,
> > >
> > > I checked A250120 by looking at your drawing. That's a cool drawing,
> and
> > > it reminds me of the Gosper Island tiling.
> > >
> > > Just by looking at your picture, it seems like your first few numbers
> are
> > > correct.
> > >
> > > I also wrote a computer program to extend the sequence. The algorithm
> > > recursively enumerates points in counted subset by expanding around
> each
> > > currently included point using the six hexagonal generators. Then a
> filter
> > > removes any duplicate vertices and vertices belonging to the uncounted
> > > lattice with Sqrt[7] spacing.
> > >
> > > This accidentally introduced another unrecorded sequence
> > >
> > > 1, 6, 15, 30, 49, 73, 102, 135, 174, 217, 265 ...
> > >
> > > Which is just the total number of points covered. The sequence given by
> > > the first derivative is your counting sequence
> > >
> > > 1, 5, 9, 15, 19, 24, 29, 33, 39, 43, 48 ...
> > >
> > > The sequence given by the second derivative is another unrecorded
> sequence
> > >
> > > 4, 4, 6, 4, 5, 5, 4, 6, 4...
> > >
> > > The first sequence approximately gives the area, the second
> approximately
> > > gives the perimeter, and the third seems to be bounded above by 2 Pi.
> > > Compare this to sequence of derivatives of circular area
> > >
> > > Pi R^2, 2 Pi R, 2 Pi, 0 , 0 ...
> > >
> > > In this case there is something weird happens along the boundary, so
> there
> > > is a sequence for third derivative
> > >
> > >  0, 2, -2, 1, 0, -1, 2, -2 ...
> > >
> > > But it appears that this sequence will have a zero average in the limit
> > > where the number of terms N approaches infinity. Maybe this palindrome
> > > pattern continues? Up to 20 terms, your sequence A250120 appears to be
> the
> > > third integral of a periodic pattern.
> > >
> > > The code I give is probably not the best way to specify this sequence.
> It
> > > should be easier to find a recursion for the second or third
> derivatives
> > > because those sequences seem to have a finite alphabet.
> > >
> > > Thanks,
> > >
> > > Brad
> > >
> > >
> > >
> > > On Sun, Nov 23, 2014 at 12:59 PM, Neil Sloane <njasloane at gmail.com>
> wrote:
> > >
> > >> There are 11 uniform (or Archimedean) tilings in the plane.
> > >> If we take the 3^6 tiling (or net) (6 triangles around each point),
> > >> start at a lattice point, and walk outwards for 0, 1, 2, 3, 4, ...
> steps,
> > >> the number of points we reach for the first time gives the sequence
> 1, 6,
> > >> 12, 18, 24, 30, 36, 42, ...,
> > >> increasing by 6 at each step after the first.
> > >> This is sequence https://oeis.org/A008458 in the OEIS.
> > >> (In other words, it is the number of nodes at graph distance n from a
> > >> fixed
> > >> node.)
> > >>
> > >> The planar net 3.6.3.6 gives https://oeis.org/A008579, and I just
> added a
> > >> primitive drawing to the entry to illustrate the first few terms.
> This is
> > >> rather more complicated.
> > >>
> > >> Next I looked at the 3^4.6 net, and for the initial terms of
> > >> the sequence I get 1,5,9,15,19,24, by hand.
> > >> This is bothersome, because (a) it is quite irregular, and (b) it was
> not
> > >> in the OEIS!  I just added it (https://oeis.org/A250120), along with
> a
> > >> drawing showing my calculations. I have no confidence in these
> numbers -
> > >> could someone check them?
> > >>
> > >> I don't know how many of the other planar nets are in the OEIS. 3^6 is
> > >> A008458, 3^4.6 is tentatively A250120, 3^3.4^2 is A008706, 3^2.4.3.4
> = ?,
> > >> 4^4 is A008574, 3.4.6.4 is ?, 3.6.3.6 is A008579, 4.8^2 is A008576,
> 6^3 is
> > >> A008486, and the others I don't know.
> > >>
> > >> Best regards
> > >> Neil
> > >>
> > >> _______________________________________________
> > >>
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> > >>
> > >
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