A215878 Lengths of loops in the P2 Penrose tiling.
10, 20, 80, 100, 460, 620, 2780, 3700, 16660, 22220, 99980, 133300, 599860, 799820, 3599180, 4798900, 21595060, 28793420, 129570380, 172760500, 777422260, 1036563020, 4664533580, 6219378100, 27987201460, 37316268620, 167923208780, 223897611700, 1007539252660, 1343385670220, 6045235515980
Offset: 1
Examples
The smallest loop a(1)=10 corresponds to the 10 kites which form the perimeter of a regular decagon.
Links
- Jacob Aron, First gliders navigate ever-changing Penrose universe, New Scientist.
- Adam P. Goucher, Blog post about this
- Adam P. Goucher, Gliders in Cellular Automata on Penrose Tilings, Journal of Cellular Automata (2012).
- Index entries for linear recurrences with constant coefficients, signature (0,5,0,6).
Programs
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Mathematica
Table[{1,1}.MatrixPower[{{5,2},{3,0}},Floor[n/2]].{10,10Mod[n,2]},{n,0,49}] Table[-(5/7)(-6^(1/2(n-1))(9+2Sqrt[6]+(-1)^n(-9+2Sqrt[6]))+4(Cos[n Pi/2] + Sin[n Pi/2])), {n, 1, 20}] (* Benedict W. J. Irwin, Nov 01 2016 *)
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PARI
Vec(-10*x*(3*x^2+2*x+1)/((x^2+1)*(6*x^2-1)) + O(x^100)) \\ Colin Barker, May 19 2014
Formula
Recurrence relation: a(n+4) = 5*a(n+2) + 6*a(n).
G.f.: -10*x*(3*x^2+2*x+1) / ((x^2+1)*(6*x^2-1)). - Colin Barker, May 19 2014
a(n) = 3*a(n-1)+2*a(n-2) if n is odd. a(n) = 2*a(n-1)-3*a(n-2) if n is even. - R. J. Mathar, Jun 18 2014
a(n) = -5 * ( -6^((n - 1)/2) * (9 + 2*sqrt(6) + (-1)^n * (2 * sqrt(6) - 9)) + 4 * (cos(n * Pi/2) + sin(n * Pi/2)))/7. - Benedict W. J. Irwin, Nov 01 2016
Comments