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A138967
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Infinite Fibonacci word on the alphabet (1,2,3,4).
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1
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| a(n)=3 for n=3,8,21,55,..., F(2k), k>1.
a(n)=4 for n=5,13,34,89,..., F(2k+1), k>1.
Start with the infinite Fibonacci word A003849, which is
0 1 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 1 0 ... and replace
each 0 by 1,2,3 and each 1 by 1,4; the result is A138967.
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FORMULA
| 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=001950. 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
| Cf. A000201, A001950, A003849, A101864.
Sequence in context: A097744 A055445 A135560 * A035612 A199539 A089555
Adjacent sequences: A138964 A138965 A138966 * A138968 A138969 A138970
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KEYWORD
| nonn
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AUTHOR
| Clark Kimberling (ck6(AT)evansville.edu), Apr 04 2008
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