[seqfan] Re: [The Tiling List] Coordination sequences for planar nets
Brad Klee
bradklee at gmail.com
Mon Nov 24 00:53:41 CET 2014
Hi Darrah,
Thanks for the correction. If you've looked at more iterations, probably
you are right. It seems like some of your email was truncated though...
I was also thinking about constructing counterclockwise ordered
circumferences from 18 vectors:
6 1st - nearest neighbor vectors
6 2nd - nearest neighbor vectors
6 3rd - nearest neighbor vectors
>From looking at the first few iterations it seems like these vectors would
be sufficient for writing a circumference; although, it might also be
better to use more vectors that encode location within a unit-cell of the
Sqrt[7] lattice.
Sounds like you are well on your way. Interested to see what you will
find...
Brad
On Sun, Nov 23, 2014 at 5:29 PM, Darrah Chavey <chavey at beloit.edu> wrote:
> 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,
>
> Brad
>
>
>
>
> 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.
>> Phone: 732 828 6098; home page: http://NeilSloane.com
>> <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
>>>>
>>>> _______________________________________________
>>>>
>>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>>
>>>
>>>
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>>
>>
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>
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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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