

A286746


{00>null}transform of the infinite Fibonacci word A003849.


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, 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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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 AlloucheDekking 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((n1)*phi) + 2*(n1)), conjectured by Baruchel for A214971, and proved in my paper on basephi 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((3phi)*n), and one easily verifies that phi+2 and 3phi 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 betaexpansions, arXiv:1906.08437 [math.NT], 2019.


FORMULA

a(n) = floor(n/(3phi))  floor((n1)/(3phi)).  Michel Dekking, Aug 16 2019


MATHEMATICA

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 *)


CROSSREFS

Cf. A003849, A214971, A195121.
Sequence in context: A123506 A051105 A284929 * A155897 A144610 A188068
Adjacent sequences: A286743 A286744 A286745 * A286747 A286748 A286749


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, May 14 2017


STATUS

approved



