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

[Neil Sloane]

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
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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