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A119346
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Sequence of nim-values for the game in which two players alternately cut off one inch or root two inches from a piece of string of length n. Player who runs out of string loses.
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1
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0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1
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OFFSET
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0,3
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COMMENTS
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It follows from Alex Fink's remarks that (a(n)) is obtained from the sequence A276862 (removing the first 2) by mapping every 2 to 0,1 and every 3 to 0,1,2. However, the first 3 entries will be missing.
In the context of my paper "Morphic words, Beatty sequences and integer images of the Fibonacci language", this means that (a(n+3)) is obtained by decorating A006337 by the decoration delta given by delta(1) = 01, delta(2) = 012. This implies that (a(n+3)) is a morphic sequence, i.e., the letter to letter image of the fixed point of a morphism, say sigma. One obtains sigma by the 'natural' algorithm given in the "Morphic words...."-paper. In turns out that the alphabet of sigma can be chosen as {0,1,2}, and that sigma is surprisingly simple:
sigma(0) = 01, sigma(1) = 012, sigma(2) = 01.
The letter to letter map is given by the identity. In other words, if x = 010120101... is the unique fixed point of sigma, then (a(n+3)) = x. (End)
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LINKS
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FORMULA
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To get the answers, add one to sequence A003151 and then start counting from zero, but return to zero whenever you reach a member of A003151 plus one.
Added Feb 13 2020: The simplest formula is a(n) = floor(n mod (1 + sqrt 2)). - Alex Fink (see link).
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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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