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A287402 Positions of 1 in A287372; complement of A287527.

%I A287402 #25 Apr 23 2018 09:33:47
%S A287402 4,8,12,19,23,31,35,43,50,54,58,65,69,77,84,91,95,103,107,115,122,129,
%T A287402 133,141,148,152,156,163,167,175,179,187,194,198,202,209,213,221,228,
%U A287402 235,239,247,254,258,262,269,276,280,284,291,295,303,310,317,321
%N A287402 Positions of 1 in A287372; complement of A287527.
%C A287402 Conjecture: a(n)/n -> 5.89..., as n -> infinity, and if m denotes this number, then -1 < m - a(n)/n <= m - 4 < 2 for n >= 1.
%C A287402 From _Michel Dekking_, Mar 18 2018: (Start)
%C A287402 Here is a proof of part of this conjecture. We recall from the comments of A287372 that A287372 = delta(x), where x is the fixed point of sigma^2 with x(1)=3. Here sigma is the morphism on {1,2,3} given by
%C A287402     sigma(1) = 2,   sigma(2) = 3,   sigma(3) = 2112,
%C A287402 and delta is the 'decoration' morphism defined by
%C A287402     delta(1) = 00,  delta(2) = 1000,  delta(3) = 0001000.
%C A287402 Let M be the incidence matrix of the morphism sigma, i.e., M equals
%C A287402     |0 0 2|
%C A287402     |1 0 2|
%C A287402     |0 1 0|.
%C A287402 The characteristic polynomial of M is equal to chi(u) = u^3-2u-2. It is well known that the frequencies mu[1], mu[2] and mu[3] in x exist, and can be computed from the Perron Frobenius eigenvalue LPF of M.
%C A287402 Solving chi(u) = 0, one finds that
%C A287402      LPF = (1/3)*(27+3*sqrt(57))^(1/3)+2/(27+3*sqrt(57))^(1/3).
%C A287402 For the frequencies one computes
%C A287402     mu[1] = 2/D,  mu[2] = LPF^2/D,  and mu[3] = LPF/D,
%C A287402 where D = LPF^2+LPF+2.
%C A287402 From the existence of these frequencies one can deduce the existence of the limit m of a(n)/n as n tends to infinity.
%C A287402 To find the value of m, note that there are
%C A287402       A(n):= N(2)(sigma^n(1)) + N(3)(sigma^n(1))
%C A287402 letters 1 in SR^n(00) = delta(sigma^n(1)), where N(i)(w) denotes the number of occurrences of the letter i in a word w.
%C A287402 The position of the A(n)-th 1 in SR^n(00) is equal to the length of SR^n(00), with an error of at most 7 positions. It follows that
%C A287402     A(n)/|SR^n(00)| -> m as n->infinity,
%C A287402 where |SR^n(00)| denotes the length of SR^n(00).
%C A287402 But
%C A287402   |SR^n(00)| = 2N(1)(sigma^n(1)) + 4N(2)(sigma^n(1)) +7N(3)(sigma^n(1)).
%C A287402 It follows therefore that
%C A287402     m = (mu[1]+mu[3])/(2mu[1]+4mu[2]+7mu[3]) =  5.899687789...
%C A287402 (End)
%H A287402 Clark Kimberling, <a href="/A287402/b287402.txt">Table of n, a(n) for n = 1..10000</a>
%t A287402 s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
%t A287402 w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "000"}]
%t A287402 Table[w[n], {n, 0, 8}]
%t A287402 st = ToCharacterCode[w[22]] - 48   (* A287372 *)
%t A287402 Flatten[Position[st, 0]]  (* A287527 *)
%t A287402 Flatten[Position[st, 1]]  (* A287402 *)
%Y A287402 Cf. A287372, A287527.
%K A287402 nonn,easy
%O A287402 1,1
%A A287402 _Clark Kimberling_, Jun 17 2017