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A289037 Positions of 1 in A289035; complement of A289036.

%I A289037 #13 Mar 22 2022 18:47:30
%S A289037 3,6,9,13,16,20,23,26,30,33,37,40,43,47,50,54,57,61,64,67,71,74,78,81,
%T A289037 84,88,91,95,98,102,105,108,112,115,119,122,125,129,132,136,139,142,
%U A289037 146,149,153,156,160,163,166,170,173,177,180,183,187,190,194,197
%N A289037 Positions of 1 in A289035; complement of A289036.
%C A289037 Conjecture:  a(n)/n -> 2 + sqrt(2), and 0 < 2 + sqrt(2) - a(n)/n < -1 + sqrt(2) for n >= 1.
%C A289037 From _Michel Dekking_, Mar 20 2022: (Start)
%C A289037 Proof of (a part of) Kimberling's conjecture. This is based on the formula for the recursion in the FORMULA section of A293077, where we use that the final-letter-removed version of the String-Replace map has the same fixed point A289035.
%C A289037 Let N1(n) be the number of 1's in the n-th iterate theta(n) of the final-letter-removed mapping defined in A289035. It was observed in A293077 that  N1(n) = L(n-1)/2, where L(n) = A293077(n) is the length of the n-th iterate theta(n) of the final-letter-removed mapping. As usual, proving a(n)/n -> 2 + sqrt(2) is the same as proving that the frequency of 1's in A293077 is equal to 1/(2 + sqrt(2)) = 1 - sqrt(2)/2 =: mu. Ignoring a proof of the existence of the limit, this means that we have to show that
%C A289037       N1(n) /L(n) -> 1 - sqrt(2)/2   as n->oo.
%C A289037 By the observation above this is the same as proving that
%C A289037       L(n-1)/2L(n) -> 1 - sqrt(2)/2   as n->oo.
%C A289037 We know from A293077 that
%C A289037       L(n) = 2 L(n-1) - L(n-2) + 2 floor(L(n-2)/4).
%C A289037 For the determination of the limit we replace the term 2 floor(L(n-2)/4) by  L(n-2)/2, as we may.
%C A289037 So we have L(n) ~ 2 L(n-1) - L(n-2)/2, which leads to
%C A289037       L(n)/2L(n+1) ~ 2 L(n-1)/2L(n+1) - L(n-2)/4L(n+1).
%C A289037 Replacing L(n-1)/L(n+1) by L(n-1)/2L(n)*2L(n)/L(n+1), and similarly for the last term, we obtain an equation for mu as n->oo:
%C A289037       mu = 4mu^2 - 2mu^3.
%C A289037 The solutions to this equation are 0, 1+sqrt(2)/2, and 1-sqrt(2)/2. Since 0<mu<1, it follows that mu = 1 - sqrt(2)/2.
%C A289037 Note: The full conjecture of Kimberling would follow from the conjecture A289035(n) = A171588(n-3) for n>3. (End)
%C A289037 Conjecture:  a(n+1) - a(n) = A276864(n) for all n. - _Michel Dekking_, Mar 20 2022
%C A289037 From _Michel Dekking_, Mar 22 2022: (Start)
%C A289037 The arguments given in the proof above carry over to the positions of 1 in the sequences A288997, A289001, A289011, A289025, A289035, A289071, A289239, and A289242. It follows that for all these sequences x(n)/n -> 2 + sqrt(2), where x(n) is the position of the n-th 1 in these sequences.
%C A289037 Note that this does not apply to the sequence A289165 where the String-Replace map is not equal to a two-block map because the 2-block 11 occurs at position 85 in A289165. (End)
%H A289037 Clark Kimberling, <a href="/A289037/b289037.txt">Table of n, a(n) for n = 1..10000</a>
%t A289037 z = 10; (* number of iterates *)
%t A289037 s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
%t A289037 w[n_] := StringReplace[w[n - 1], {"00" -> "0010", "01" -> "010", "10" -> "010"}]
%t A289037 TableForm[Table[w[n], {n, 0, 10}]]
%t A289037 st = ToCharacterCode[w[z]] - 48   (* A289035 *)
%t A289037 Flatten[Position[st, 0]]  (* A289036 *)
%t A289037 Flatten[Position[st, 1]]  (* A289037 *)
%Y A289037 Cf. A289035, A289036.
%K A289037 nonn,easy
%O A289037 1,1
%A A289037 _Clark Kimberling_, Jun 27 2017