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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 03 2017
STATUS
approved