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A308590
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Langton's ant on a Penrose rhomb tiling: number of black cells after n moves of the ant.
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11
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0, 1, 2, 3, 4, 5, 4, 5, 6, 7, 8, 9, 8, 7, 6, 7, 6, 7, 8, 9, 8, 9, 10, 11, 10, 9, 8, 7, 8, 7, 8, 9, 10, 9, 10, 11, 12, 11, 12, 13, 14, 13, 14, 15, 16, 15, 16, 17, 18, 17, 16, 15, 14, 13, 12, 11, 12, 11, 12, 13, 14, 13, 14, 13, 14, 15, 14, 13, 12, 13, 12, 13, 14
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OFFSET
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0,3
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COMMENTS
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The ant lives on a centrally symmetric Penrose rhomb tiling with a "Sun" patch (S configuration, cf. A242935) at the center and starts on one of the thick rhombs of that patch, looking towards one of the outward edges of that tile. On a white rhomb, turn to the next edge of that cell in clockwise direction, flip the color of the rhomb, then move forward one unit. On a black rhomb, turn to the next edge of that cell in counterclockwise direction, flip the color of the rhomb, then move forward one unit.
In contrast to the corresponding sequences for Langton's ant on periodic tilings, like the square tiling (A255938) or a hexagonal tiling (A269757), this sequence is most likely not unique. A Penrose tiling lacks translational symmetry, meaning any two finite regions in the tiling that are identical are surrounded by different patches of tiles when examining a large enough region of the surrounding tiles. Therefore I suspect that, unless the trajectory of the ant is bounded to stay inside a finite region of the tiling, the trajectories of any two ants placed at different starting points on the tiling will diverge at some point.
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LINKS
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EXAMPLE
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See illustration in links.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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