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%I #19 Dec 28 2017 21:35:48
%S 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,
%T 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,
%U 1,1,0,0,1,0,0,1,1,0,0,1,1,0,0,1,0,0
%N {0->1, 1->00}-transform of the infinite Fibonacci word A003849.
%C Conjecture: the positions of 1 are given by A050140. This has been checked for the first million terms.
%C From _Michel Dekking_, Dec 28 2017: (Start)
%C Proof of the conjecture:
%C 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.
%C It suffices to prove that b(n+1)-b(n) = d(n), where b is the sequence of positions of 1 in a = A287772.
%C 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)
%H Clark Kimberling, <a href="/A287772/b287772.txt">Table of n, a(n) for n = 1..10000</a>
%e As a word, A003849 = 0100101001001010010100100..., and replacing each 0 by 1 and each 1 by 00 gives 1001100100110011001001100100110011001...
%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)
%t w = StringJoin[Map[ToString, s]]
%t w1 = StringReplace[w, {"0" -> "1", "1" -> "00"}]
%t st = ToCharacterCode[w1] - 48 (* A287772 *)
%t Flatten[Position[st, 0]] (* A287775 *)
%t Flatten[Position[st, 1]] (* A050140 conjectured *)
%Y Cf. A050140, A050141, A287775.
%K nonn,easy
%O 1
%A _Clark Kimberling_, Jun 03 2017
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