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%I #14 Dec 28 2017 21:36:01
%S 2,3,6,7,9,10,13,14,17,18,20,21,24,25,27,28,31,32,35,36,38,39,42,43,
%T 46,47,49,50,53,54,56,57,60,61,64,65,67,68,71,72,74,75,78,79,82,83,85,
%U 86,89,90,93,94,96,97,100,101,103,104,107,108,111,112,114
%N Positions of 0 in A287772; complement of A050140 (conjectured and proved).
%C -1 < n*r - a(n) < 1 for n >= 1, where r = (5 + sqrt(5))/4.
%C From _Michel Dekking_, Dec 28 2017: (Start)
%C Let (d(n)) be the sequence of first differences: d(n)=a(n+1)-a(n).
%C CLAIM: d(n) = A108103(n+1) for n=1,2,….
%C Proof: As a word A287772 = 100110010011001100100110010011... obtained by substituting 0->1, 1->00 in the Fibonacci word F=0100101001001010010100...
%C This implies that A287772 is a concatenation of 00’s separated by 1’s and 11’s. Moreover, a 0110 occurs iff 1001 occurs in F, and a 010 occurs iff 101 occurs in F. Note also that occurrence of a 00 in A287772 yields a d(n)=1 (and so every other letter in d is a 1), occurrence of a 010 yields a d(n)=2, and occurrence of a 0110 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)=A108103(n+1). (End)
%H Clark Kimberling, <a href="/A287775/b287775.txt">Table of n, a(n) for n = 1..10000</a>
%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, A287772, A108103.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Jun 03 2017
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