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Number of periodic sequences of period 3n generated by the random period doubling substitution 0 --> {01, 10}, 1 --> {00}.
0

%I #11 Dec 30 2019 00:13:07

%S 3,15,21,375,108,2427,402,176391,1533,216030,10992,19375935,24612,

%T 13106514

%N Number of periodic sequences of period 3n generated by the random period doubling substitution 0 --> {01, 10}, 1 --> {00}.

%C From an initial seed letter 0, the random substitution iteratively acts on words and outputs all possible outcomes from applying all combinations of the allowed substitution rules independently to each letter in a word. So we have:

%C 0 -->

%C {01, 10} -->

%C {0100, 1000, 0001, 0010} -->

%C {01000101, 01000110, 01001001, 01001010, 10000101, 10000110, 10001001, 10001010, 00010101, 00010110, 00011001, 00011010, 00100101, 00100110, 00101001, 00101010, 01010100, 01011000, 01100100, 01101000, 10010100, 10011000, 10100100, 10101000, 01010001, 01010010, 01100001, 01100010, 10010001, 10010010, 10100001, 10100010} --> ...

%C An infinite sequence is generated by the random substitution if all subwords of the sequence appear as subwords of some word appearing in the infinite list generated above. Some of these infinite sequences will be periodic and so we can enumerate them.

%C All periodic sequences have period a multiple of 3.

%H D. Rust, <a href="https://www.math.uni-bielefeld.de/~drust/papers/random_subs_periodic_points.pdf">Periodic points in random substitution subshifts</a>, (2018).

%e The periodic sequences of length 3 generated by the random period doubling substitution have periodic blocks 001, 010, 100.

%e The periodic sequences of length 6 generated by the random period doubling substitution have periodic blocks 010100, 101000, 010001, 100010, 000101, 001010, 011000, 110000, 100001, 000011, 000110, 001100, 100100, 001001, 010010.

%Y Cf. A096268, A275202.

%K nonn,more

%O 1,1

%A _Dan Rust_, Aug 18 2018