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A138967
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Infinite Fibonacci word on the alphabet {1,2,3,4}.
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2
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1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3
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OFFSET
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1,2
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COMMENTS
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Start with the infinite Fibonacci word A003849, which is 0100101001001010010... and replace each 0 by 1,2,3 and each 1 by 1,4.
(a(n)) is the unique fixed point of the morphism 1->12, 2->3, 3->14, 4->3, obtained by coding the overlapping 3-block morphism of the Fibonacci morphism according to 010<->1, 100<->2, 001<->3, 101<->4. - Michel Dekking, Sep 28 2017
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LINKS
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FORMULA
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a(n) = 3 for n = 3, 8, 21, 55, ..., F(2*k), where k>1.
a(n) = 4 for n = 5, 13, 34, 89, ..., F(2*k+1), where k>1.
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. a(n)=1 if n=A(A(k)) for some k; a(n)=2 if n=B(A(k)) for some k; a(n)=3 if n=A(B(k)) for some k; a(n)=4 if n=B(B(k)) for some k.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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