A287772 {0->1, 1->00}-transform of the infinite Fibonacci word A003849.

1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0

Offset

1

Comments

Conjecture: the positions of 1 are given by A050140. This has been checked for the first million terms.

From Michel Dekking, Dec 28 2017: (Start)

Proof of the conjecture:

Let F = A003849 be the Fibonacci word, and let (d(n)) = A050141 = 3,1,3,3,1,3,1,3,3,.. be the sequence of first differences of A050140.

It suffices to prove that b(n+1)-b(n) = d(n), where b is the sequence of positions of 1 in a = A287772.

Note that A287772 is a concatenation of 00’s separated by 1’s and 11’s, since 11 does not occur in F. Moreover, a 0110 occurs iff 1001 occurs in F, and a 010 occurs iff 101 occurs in F. Note also that occurrence of a 0110 in A287772 yields a d(n)=1, and occurrence of a 010 yields a d(n)=3. Since the 1001’s and 101’s occur in 1F according to F itself with 1 prepended (see A001468 and A282162), we must have d(n)= A050141. (End)

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

Example

As a word, A003849 = 0100101001001010010100100..., and replacing each 0 by 1 and each 1 by 00 gives 1001100100110011001001100100110011001...

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "1", "1" -> "00"}]
st = ToCharacterCode[w1] - 48    (* A287772 *)
Flatten[Position[st, 0]]  (* A287775 *)
Flatten[Position[st, 1]]  (* A050140 conjectured *)

Crossrefs

Cf. A050140, A050141, A287775.

Sequence in context: A120527 A188093 A190843 * A373791 A190198 A071004

Adjacent sequences: A287769 A287770 A287771 * A287773 A287774 A287775

Keyword

nonn,easy

Author

Clark Kimberling, Jun 03 2017

Status

approved