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 A289037 Positions of 1 in A289035; complement of A289036. 5
 3, 6, 9, 13, 16, 20, 23, 26, 30, 33, 37, 40, 43, 47, 50, 54, 57, 61, 64, 67, 71, 74, 78, 81, 84, 88, 91, 95, 98, 102, 105, 108, 112, 115, 119, 122, 125, 129, 132, 136, 139, 142, 146, 149, 153, 156, 160, 163, 166, 170, 173, 177, 180, 183, 187, 190, 194, 197 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Conjecture: a(n)/n -> 2 + sqrt(2), and 0 < 2 + sqrt(2) - a(n)/n < -1 + sqrt(2) for n >= 1. From Michel Dekking, Mar 20 2022: (Start) 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. 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 N1(n) /L(n) -> 1 - sqrt(2)/2 as n->oo. By the observation above this is the same as proving that L(n-1)/2L(n) -> 1 - sqrt(2)/2 as n->oo. We know from A293077 that L(n) = 2 L(n-1) - L(n-2) + 2 floor(L(n-2)/4). For the determination of the limit we replace the term 2 floor(L(n-2)/4) by L(n-2)/2, as we may. So we have L(n) ~ 2 L(n-1) - L(n-2)/2, which leads to L(n)/2L(n+1) ~ 2 L(n-1)/2L(n+1) - L(n-2)/4L(n+1). 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: mu = 4mu^2 - 2mu^3. The solutions to this equation are 0, 1+sqrt(2)/2, and 1-sqrt(2)/2. Since 03. (End) Conjecture: a(n+1) - a(n) = A276864(n) for all n. - Michel Dekking, Mar 20 2022 From Michel Dekking, Mar 22 2022: (Start) 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. 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) LINKS Clark Kimberling, Table of n, a(n) for n = 1..10000 MATHEMATICA z = 10; (* number of iterates *) s = {0, 0}; w[0] = StringJoin[Map[ToString, s]]; w[n_] := StringReplace[w[n - 1], {"00" -> "0010", "01" -> "010", "10" -> "010"}] TableForm[Table[w[n], {n, 0, 10}]] st = ToCharacterCode[w[z]] - 48 (* A289035 *) Flatten[Position[st, 0]] (* A289036 *) Flatten[Position[st, 1]] (* A289037 *) CROSSREFS Cf. A289035, A289036. Sequence in context: A066343 A172262 A184909 * A060605 A325228 A278449 Adjacent sequences: A289034 A289035 A289036 * A289038 A289039 A289040 KEYWORD nonn,easy AUTHOR Clark Kimberling, Jun 27 2017 STATUS approved

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Last modified August 6 18:55 EDT 2024. Contains 374981 sequences. (Running on oeis4.)