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 A309075 Total number of black cells after n iterations of Langton's ant with two ants on the grid placed side-by-side on neighboring squares and initially looking in the same direction. 1

%I #12 Jul 25 2019 08:31:18

%S 0,2,2,4,6,6,8,8,8,6,6,4,2,2,0,2,2,4,6,6,8,8,8,6,6,4,2,2,0,2,2,4,6,6,

%T 8,8,8,6,6,4,2,2,0,2,2,4,6,6,8,8,8,6,6,4,2,2,0,2,2,4,6,6,8,8,8,6,6,4,

%U 2,2,0,2,2,4,6,6,8,8,8,6,6,4,2,2,0,2,2

%N Total number of black cells after n iterations of Langton's ant with two ants on the grid placed side-by-side on neighboring squares and initially looking in the same direction.

%C Periodic with period 14.

%C The two ants are caught in a repeating cycle where they build and then erase a pattern of black cells, alternating between facing "northwards" and "southwards" on the completely white grid.

%H Felix Fröhlich, <a href="/A309075/a309075.pdf">Illustration of iterations 0-14 of the ants</a>, 2019.

%F Conjectures from _Colin Barker_, Jul 11 2019: (Start)

%F G.f.: 2*x*(1 + x)*(1 - x + x^2)*(1 + x^2)^2*(1 + x^4) / ((1 - x)*(1 - x + x^2 - x^3 + x^4 - x^5 + x^6)*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)).

%F a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6) + a(n-7) - a(n-8) + a(n-9) - a(n-10) + a(n-11) - a(n-12) + a(n-13) for n>12.

%F (End)

%e See illustrations in Fröhlich, 2019.

%Y Cf. A326352.

%K nonn

%O 0,2

%A _Felix Fröhlich_, Jul 10 2019

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