

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
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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 * A190198 A071004 A188083
Adjacent sequences: A287769 A287770 A287771 * A287773 A287774 A287775


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Jun 03 2017


STATUS

approved



