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%I #23 Dec 25 2018 17:28:53
%S 2,2,1,0,2,1,0,2,1,2,0,1,0,0,2,0,0,1,2,0,1,0,0,1,0,0,1,2,2,2,2,2,1,2,
%T 2,1,2,2,2,2,2,2,2,2,1,2,2,2,2,2,1,2,2,1,2,2,2,2,2,1,2,2,2,2,2,2,2,2,
%U 2,2,2,1,2,2,2,2,2,1,2,2,1,0,2,1,0,2,1
%N a(n) is the period of cyclic structures that appear in the 3-state (0,1,2) 1D cellular automaton started from a single cell at state 1 with rule n.
%C The length of the sequence is equal to 3^3^3 = 7625597484987.
%e 1D cellular automaton with rule=1 gives the following generations:
%e 1 ..........1.......... <------ start
%e 2 111111111...111111111 <------ end
%e 3 ..........1..........
%e 4 111111111...111111111
%e 5 ..........1..........
%e 6 111111111...111111111
%e 7 ..........1..........
%e The period is 2, thus a(1) = 2.
%e For rule=150:
%e 1 ..........1..... <------ start
%e 2 .........22..... <------ end
%e 3 ........1.......
%e 4 .......22.......
%e 5 ......1.........
%e 6 .....22.........
%e 7 ....1...........
%e The period is 2, thus a(150) = 2.
%e For rule=100000000797:
%e 1 .........1....... <------ start
%e 2 ........2.2......
%e 3 ........111......
%e 4 .......2.112.....
%e 5 .......12........
%e 6 ......21.........
%e 7 ........2........ <------ end
%e 8 ........1........
%e 9 .......2.2.......
%e 10 .......111.......
%e 11 ......2.112......
%e 12 ......12.........
%e 13 .....21..........
%e 14 .......2.........
%e 15 .......1.........
%e The period is 7, thus a(100000000797) = 7.
%e a(10032729) = 12.
%e a(10096524) = 16.
%t Table[
%t Length[
%t Last[
%t FindTransientRepeat[(Internal`DeleteTrailingZeros[
%t Reverse[Internal`DeleteTrailingZeros[#]]]) & /@
%t CellularAutomaton[{i, 3}, {ConstantArray[0, 25], {1}, ConstantArray[0, 25]} // Flatten, 50], 2]]],
%t {i, 1, 1000}
%t ]
%Y Cf. A180001.
%K nonn,fini
%O 1,1
%A _Philipp O. Tsvetkov_, Sep 27 2018