A287072 Start with 0 and repeatedly substitute 0->01, 1->21, 2->0.
0, 1, 2, 1, 0, 2, 1, 0, 1, 0, 2, 1, 0, 1, 2, 1, 0, 1, 0, 2, 1, 0, 1, 2, 1, 0, 2, 1, 0, 1, 2, 1, 0, 1, 0, 2, 1, 0, 1, 2, 1, 0, 2, 1, 0, 1, 0, 2, 1, 0, 1, 2, 1, 0, 2, 1, 0, 1, 2, 1, 0, 1, 0, 2, 1, 0, 1, 2, 1, 0, 2, 1, 0, 1, 0, 2, 1, 0, 1, 2, 1, 0, 1, 0, 2, 1
Offset: 1
Links
- Clark Kimberling, Table of n, a(n) for n = 1..10000
- James Currie, Pascal Ochem, Narad Rampersad, and Jeffrey Shallit, Properties of a Ternary Infinite Word, arXiv:2206.01776 [cs.DM], 2022.
- James Currie, Pascal Ochem, Narad Rampersad, and Jeffrey Shallit, Complement Avoidance in Binary Words, arXiv:2209.09598 [math.CO], 2022.
- Index entries for sequences that are fixed points of mappings
Programs
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Mathematica
s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {2, 1}, 2 -> 0}] &, {0}, 10] (* A287072 *) Flatten[Position[s, 0]] (* A287073 *) Flatten[Position[s, 1]] (* A287074 *) Flatten[Position[s, 2]] (* A287075 *) SubstitutionSystem[{0->{0,1},1->{2,1},2->{0}},{0},{8}][[1]] (* Harvey P. Dale, Feb 18 2025 *)
Comments