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A270788
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Unique fixed point of the 3-symbol Fibonacci morphism phi-hat_2.
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13
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1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3
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OFFSET
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1,2
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COMMENTS
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Fixed point of the morphism phi-hat_2 given by 1 --> 12, 2 --> 3, 3 --> 12. [Joerg Arndt, Apr 10 2016]
This sequence is the [0->12, 1->3]-transform of the Fibonacci word A003849: if T(0):=12, T(1):=3, then one proves easily with induction that T(phi_1^n(0)) = phi-hat_2^{n+1}(1), and T(phi_1^n(1)) = phi-hat_2^{n+1}(2), where phi_1 denotes the Fibonacci morphism given by 0 --> 01, 1 --> 0. - Michel Dekking, Dec 29 2019
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LINKS
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FORMULA
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Let A(n)=floor(n*tau), B(n)=n+floor(n*tau), i.e., A and B are the lower and upper Wythoff sequences, A=A000201, B=A001950. Then a(n)=1 if n=A(A(k)) for some k; a(n)=2 if n=B(k) for some k; a(n)=3 if n=A(B(k)) for some k. - Michel Dekking, Dec 27 2016
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MAPLE
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with(ListTools);
psi:=proc(S)
Flatten(subs( {1=[1, 2], 2=[3], 3=[1, 2]}, S));
end;
S:=[1];
for n from 1 to 10 do S:=psi(S): od:
S;
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MATHEMATICA
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m = 121; (* number of terms required *)
S[1] = {1};
S[n_] := S[n] = SubstitutionSystem[{1 -> {1, 2}, 2 -> {3}, 3 -> {1, 2}}, S[n-1]];
For[n = 2, True, n++, If[PadRight[S[n], m] == PadRight[S[n-1], m], Print["n = ", n]; Break[]]];
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CROSSREFS
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Cf. A159917 (same sequence if we map 1->2, 2->0, 3->1).
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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