OFFSET
1
COMMENTS
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.
Is this A230603 with an offset changed by 2? - R. J. Mathar, May 25 2017
From Michel Dekking, Aug 16 2019: (Start)
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
REFERENCES
J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
J.-P. Allouche, F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424 [math.NT], 2018.
M. Dekking, Base phi representations and golden mean beta-expansions, arXiv:1906.08437 [math.NT], 2019.
FORMULA
a(n) = floor(n/(3-phi)) - floor((n-1)/(3-phi)). - Michel Dekking, Aug 16 2019
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 14 2017
STATUS
approved