A285392 Start with a single cell at coordinates (0, 0), then iteratively subdivide the grid into 3 X 3 cells and remove the cells whose sum of modulo 2 coordinates is 0; a(n) is the number of cells after n iterations.
1, 5, 36, 264, 1944, 14328, 105624, 778680, 5740632, 42321528, 312006168, 2300197176, 16957700568, 125016939000, 921660044184, 6794737129656, 50092713636696, 369297577174392, 2722565630929176, 20071519752269880, 147972890199278808, 1090897774766270712
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Peter Karpov, InvMem, Item 26
- Peter Karpov, Illustration of cell configuration after 5 iterations
- Index entries for linear recurrences with constant coefficients, signature (9,-12).
Programs
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Magma
I:=[5, 36]; [1] cat [n le 2 select I[n] else 9*Self(n-1) - 12*Self(n-2): n in [1..31]]; // G. C. Greubel, Dec 11 2021
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Mathematica
{1}~Join~LinearRecurrence[{9, -12}, {5, 36}, 16] -
PARI
Vec((1 - x)*(1 - 3*x) / (1 - 9*x + 12*x^2) + O(x^30)) \\ Colin Barker, Apr 18 2017
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Sage
[(1/4)*(bool(n==0) + (2*sqrt(3))^(n-1)*( 6*sqrt(3)*chebyshev_U(n, 9/(4*sqrt(3))) - 7*chebyshev_U(n-1, 9/(4*sqrt(3))) ) ) for n in (0..30)] # G. C. Greubel, Dec 11 2021
Formula
a(0) = 1, a(1) = 5, a(2) = 36, a(n) = 9*a(n-1) - 12*a(n-2).
G.f.: (1-4*x+3*x^2)/(1-9*x+12*x^2).
a(n) = (2^(-3-n)*((9-sqrt(33))^n*(-13+3*sqrt(33)) + (9+sqrt(33))^n*(13+3*sqrt(33)))) / sqrt(33) for n>0. - Colin Barker, Apr 18 2017
a(n) = (1/4)*([n=0] + (2*sqrt(3))^(n-1)*( 6*sqrt(3)*ChebyshevU(n, 9/(4*sqrt(3))) - 7*ChebyshevU(n-1, 9/(4*sqrt(3))) ) ). - G. C. Greubel, Dec 11 2021
Extensions
More terms from Colin Barker, Apr 18 2017
Comments