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%I #21 Sep 28 2017 09:00:43
%S 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,
%T 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,
%U 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
%N Infinite Fibonacci word on the alphabet {1,2,3,4}.
%C Start with the infinite Fibonacci word A003849, which is 0100101001001010010... and replace each 0 by 1,2,3 and each 1 by 1,4.
%C (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
%H F. Michel Dekking, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Dekking/dekk4.html">Morphisms, Symbolic Sequences, and Their Standard Forms</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
%F a(n) = 3 for n = 3, 8, 21, 55, ..., F(2*k), where k>1.
%F a(n) = 4 for n = 5, 13, 34, 89, ..., F(2*k+1), where k>1.
%F 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.
%Y Cf. A000201, A001950, A003849, A101864, A270788, A276757.
%K nonn
%O 1,2
%A _Clark Kimberling_, Apr 04 2008
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