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A286746
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{00->null}-transform of the infinite Fibonacci word A003849.
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1
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0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1
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OFFSET
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1
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COMMENTS
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As a word, A003849 = 01001010010010100101001001010010..., and deleting each occurrence of 00 gives 01101110110111011101101110110111011..., in which, it is conjectured, the positions of 0 are given by A214971, and of 1, by A195121.
Proof of the two conjectures by Kimberling: first note that the [00->null]-transform is the same as the [001->1]-transform, since 000 does not occur in the infinite Fibonacci word.
Next, we do the following trick: replace the [001->1]-transform by the [001->2]-transform. Then the Fibonacci word is mapped to b:= A284749 = 0120122012... Note that the positions of 0 in b are the same as the positions of 0 in a. By Theorem 31 in the Allouche-Dekking paper, the positions of 0 in b are given by the sequence with terms floor(n*phi)+2*n+1, for n=0,1,2.... Transforming to offset 1, this is the sequence (ceiling((n-1)*phi) + 2*(n-1)), conjectured by Baruchel for A214971, and proved in my paper on base-phi representations.
The positions of 1 in (a(n)) are given by A195121 for n > 0, since by the Comments in A195121 this sequence has terms 3*n - floor(n*phi) = floor((3-phi)*n), and one easily verifies that phi+2 and 3-phi form a Beatty pair.
(End)
Proof of the conjecture by Mathar: this follows directly from Lemma 9.1.3 in the book by Allouche and Shallit. - Michel Dekking, Aug 16 2019
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REFERENCES
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J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
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LINKS
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FORMULA
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a(n) = floor(n/(3-phi)) - floor((n-1)/(3-phi)). - Michel Dekking, Aug 16 2019
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 12]; (* A003849 *)
w = StringJoin[Map[ToString, s]];
w1 = StringReplace[w, {"00" -> ""}]; st = ToCharacterCode[w1] - 48; (* A286746 *)
Flatten[Position[st, 0]]; (* A214971 *)
Flatten[Position[st, 1]]; (* A195121 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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