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

Darrah Chavey chavey at beloit.edu
Mon Nov 24 00:29:19 CET 2014

```One of the interesting structural elements of the vertices at distance d from the center is that when we connect them by the edges of the tiling, then they fall into C(d) connected components, and when d > 5, C(d) = C(d–5) + 6. This is one geometric correlation with the length 5 periodicity of the coordination sequence. It also means that Brad's suspicion that "there will be a finite number of circumference-segments" is not correct -- assuming that I have correctly interpreted what he means by "circumference segments".

the number of connected components
On Nov 23, 2014, at 5:13 PM, Brad Klee <bradklee at gmail.com> wrote:

> Hi Neil,
>
> You are welcome to use the image I made, or the new image attached here.
>
> I added some lines to the background, which makes the image easier to use for exploration and better quality.
>
> ~~
>
> I'm not sure what you mean about the generating polynomial or PORC? How does the polynomial generating function lead to a proof of periodicity or better understanding of the structure of the tiling / sequence?
>
> The simple geometric explanation via area, circumference, etc. would be the only way for me to find out more about the tiling / sequence.
>
> Notice that each element in the series corresponds to a topological circle. Connecting each nearest point into a cycle, the approximately-circular curves do not appear to cross. For each curve there exists another partitioning up to cyclic permutations:
>
> 9:     { 2, 3, 4 }
> 15:   { 6, 2, 7 }
> 19:   { 2, 2, 6, 4, 5 }
> 24:   { 2, 5, 2, 4, 5, 6 }
> ...
>
> which is obtained by observing that each consecutive point along any circumference is separated by a distance 1, Sqrt[3], or 2 (conjectured), and counting the number of consecutive points joined by distances of 1.
>
> My intuition tells me that there will be a finite number of circumference-segments, and each of these can be uniquely identified with a finite alphabet of symbols { A , B, C, D, E, F, ... }. Under iteration, the next set of symbols along the circumference ( n + 1 ) is entirely determined by pairs of symbols along the circumference ( n ).
>
> For example, the Blue -> Green replacement is given by
>
>  2 , 7 ---> 6
>  7 ---> 4
> 7 , 6  ---> 5
> 6 ---> 2, 2
>
> Using letters rather than numbers, this becomes
>
>  A , B ---> C****
>  B ---> D
> B , C  ---> E
> C ---> F, A*
>
> where star indicates rotation by 2 Pi / 6 radians. Notice that F & A plus rotations are different due to their inequivalent occurrence with regard to the Sqrt[7] lattice. A better encoding would require explicit encoding that separations can be either 2 or Sqrt[3].
>
> If all of my assumptions about finite alphabets are valid, this approach of expanding in circumference fragments could even lead to a induction proof of periodicity. Though it's difficult to say what will happen with a replacement system on an alphabet of symbols probably order of magnitude 100 or 1000.
>
> Another idea is to define a set of annuli such that the union of all annuli is the plane and each annuli contains only points of a given circumference. Perhaps a smart definition of the annuli could then be used to count the points.
>
> Thanks,
>
>
>
>
>
> On Sun, Nov 23, 2014 at 2:31 PM, Neil Sloane <njasloane at gmail.com> wrote:
> Brad, I was just about to reply when Maurizio's message arrived. All the terms that you and he found can be explained
> by the (conjectured) generating function
> (x^2+x+1)*(x^4+3*x^3+3*x+1)/((x^4+x^3+x^2+x+1)*(x-1)^2),
> which is nice and symmetric.
>
> Finding a proof should be easy, now we know the answer.
>
> I'll update the entry (A250120) later today.
> And I'll certainly include your picture, which
> is far nicer than mine. (I assume that's OK?)
> I'll include both pictures, yours and mine.
>
> 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.
> 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,
>
>
>
>
> 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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> <IntegerSequence.png><IntegerSequence.nb>

Darrah Chavey
Professor, Math & Computer Science
Beloit College, Wisc.
http://cs.beloit.edu/chavey/

I wrote a novel about a fellow who had a small garden. It didn't have much of a plot.

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