[seqfan] Re: Coordination sequences for planar nets
Brad Klee
bradklee at gmail.com
Sun Nov 23 21:04:21 CET 2014
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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