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%I #11 Aug 30 2020 03:37:19
%S 1,2,5,6,7,10,11,14,15,16,19,20,21,24,25,28,29,30,33,34,37,38,39,42,
%T 43,44,47,48,51,52,53,56,57,58,61,62,65,66,67,70,71,74,75,76,79,80,81,
%U 84,85,88,89,90,93,94,97,98,99,102,103,104,107,108,111,112
%N Positions of 1 in A286749; complement of A286750.
%C a(n) - a(n-1) is in {1,2,3} for n>=2, and a(n)/n -> 4 - sqrt(5).
%C From _Michel Dekking_, Aug 29 2020: (Start)
%C This sequence is a generalized Beatty sequence.
%C Recall from A286749 that A286749 is the letter-to-letter image of the fixed point x of the morphism mu given by
%C mu: 1->12341, 2->1, 3->2, 4->34.
%C where the letter-to-letter map lambda is defined by
%C lambda: 1->1, 2->1, 3->0, 4->0.
%C The return words of the word 1 in x are A:=1 and B:=1234.
%C We have mu(1)=12341, and mu(1234)=123411234. So the derived morphism is tau: A->BA, B->BAB.
%C This morphism happens to be the square of the Fibonacci morphism on the alphabet {B,A}.
%C The return word A = 1 has lambda-image 1, and the return word B = 1234 has lambda-image 1100. This means that they give distances 1, respectively 1 and 3 between (successive) occurrences of 1's in A286749. This leads to the decoration B->13, A->1, which amounts to replacing 0 by 01 and 1 by 0 in the Fibonacci word.
%C But the Fibonacci word is fixed by the {0,1} version of tau. It follows that the sequence (a(n+1)-a(n)) = 1,3,1,1,3,... is the Fibonacci word on the alphabet {1,3}. Finally, Lemma 8 in the paper "Generalized Beatty sequences..." then gives that a(n) = 5*n-2*floor(n*phi)-2, for n>0.
%C Clearly this implies a(n)/n -> 4 - sqrt(5).
%C (End)
%H Clark Kimberling, <a href="/A286751/b286751.txt">Table of n, a(n) for n = 1..10000</a>
%H J.-P. Allouche, F. M. Dekking, <a href="https://arxiv.org/abs/1809.03424">Generalized Beatty sequences and complementary triples</a>, arXiv:1809.03424 [math.NT], 2018-2019.
%F a(n) = 5*n-2*floor(n*phi)-2. - _Michel Dekking_, Aug 29 2020
%e As a word, A286749 = 11001110011001110011010..., in which 1 is in positions 1,2,5,6,7,10,...
%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 12]; (* A003849 *)
%t w = StringJoin[Map[ToString, s]];
%t w1 = StringReplace[w, {"0100" -> ""}]; st = ToCharacterCode[w1] - 48; (* A286749 *)
%t Flatten[Position[st, 0]]; (* A286750 *)
%t Flatten[Position[st, 1]]; (* A286751 *)
%Y Cf. A003849, A286749, A286750.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, May 14 2017
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