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Apply the morphism 1 -> {1, 2}, 2 -> {3,1}, 3 -> {1} n times to 1, and concatenate the resulting string.
2

%I #7 Aug 21 2018 10:21:35

%S 1,12,1231,1231112,1231112121231,123111212123112311231112,

%T 12311121212311231123111212311121231112121231,

%U 123111212123112311231112123111212311121212311231112121231123111212123112311231112

%N Apply the morphism 1 -> {1, 2}, 2 -> {3,1}, 3 -> {1} n times to 1, and concatenate the resulting string.

%C For the tribonacci word A092782, each block b(n) (see A103269) is the concatenation of the three previous blocks: b(n) = b(n-1).b(n-2).b(n-3). Instead, here we have a(n) = a(n-1).a(n-3).a(n-2), as can be seen in the examples section below.

%D V. F. Sirvent, Semigroups and the self-similar structure of the flipped tribonacci substitution, Applied Math. Letters, 12 (1999), 25-29.

%e a(1): 1,

%e a(2): 12,

%e a(3): 1231,

%e a(4): 1231112,

%e a(5): 1231112121231,

%e a(6): 123111212123112311231112,

%e a(7): 12311121212311231123111212311121231112121231,

%e equals a(6).a(4).a(5), look:

%e a(6):123111212123112311231112,

%e a(4): 1231112,

%e a(5): 1231112121231,

%e a(8): 123111212123112311231112123111212311121212311231112121231123111212123112311231112

%e equals a(7).a(5).a(6), look:

%e a(7): 12311121212311231123111212311121231112121231,

%e a(5): 1231112121231,

%e a(6): 123111212123112311231112,

%Y Cf. A100619 (the limiting string), A277735, A317953.

%Y A103269 is the analog for the word A092782.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Aug 21 2018