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A287772
{0->1, 1->00}-transform of the infinite Fibonacci word A003849.
4
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
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 03 2017
STATUS
approved