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%I #18 Sep 08 2022 08:46:19
%S 1,7,13,17,23,29,33,39,43,49,55,59,65,71,75,81,85,91,97,101,107,111,
%T 117,123,127,133,139,143,149,153,159,165,169,175,181,185,191,195,201,
%U 207,211,217,221,227,233,237,243,249,253,259,263,269,275,279,285,289
%N Positions of 1 in A288711; complement of A288712.
%C Conjecture: a(n)/n-> 3 + sqrt(5), and if m denotes this number, then -1 < m - a(n)/n) < 5 for n >= 1.
%C The truth of this conjecture follows directly from the same statement for A288709, in combination with Mathar's Jun 19 2017 conjecture. - _Michel Dekking_, Oct 20 2018
%C Appears to differ from A288709 only at a(1). - _R. J. Mathar_, Jun 19 2017
%C From _Michel Dekking_, Oct 20 2018: (Start)
%C Here is a proof of this conjecture.
%C The sequence (a(n)) consists of the positions of 1 in A288711, which can be obtained from A289034 by a [0->00, 1->10]-transform (see Kimberling's observation in the Comments of A288707). Similarly, the sequence A288709 consists of the positions of 1 in A288707, which can be obtained from A189661 by the same [0->00, 1->10]-transform. It is shown in the Comments of A189661 that A189661 = A289034, except for the first two terms. In A189661 these two terms are 0,1, which is transformed to 0,0,1,0. In A289034 these two terms are 1,0, which is transformed to 1,0,0,0. We see that the first 1 in A288711 occurs at position 1, but the first 1 in A288707 appears at position 3. The next 1's will occur at the same position in the two sequences.
%C Conclusion: the sequence (a(n)) only differs from A288709 at a(1). (End)
%H Clark Kimberling, <a href="/A288713/b288713.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n+1) = 2*floor(n*phi) +2*n + 3, where phi = (1+sqrt(5))/2. - _Michel Dekking_, Oct 20 2018
%t s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
%t w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "00"}]
%t Table[w[n], {n, 0, 8}]
%t st = ToCharacterCode[w[9]] - 48 (* A288711 *)
%t Flatten[Position[st, 0]] (* A288712 *)
%t Flatten[Position[st, 1]] (* A288713 *)
%o (Magma) [1] cat [2*Floor(n*(1 + Sqrt(5)) / 2) + 2*n + 3: n in [1..60]]; // _Vincenzo Librandi_, Oct 20 2018
%Y Cf. A288711, A288712.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jun 16 2017
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