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Positions of 1 in A287372; complement of A287527.
3

%I #25 Apr 23 2018 09:33:47

%S 4,8,12,19,23,31,35,43,50,54,58,65,69,77,84,91,95,103,107,115,122,129,

%T 133,141,148,152,156,163,167,175,179,187,194,198,202,209,213,221,228,

%U 235,239,247,254,258,262,269,276,280,284,291,295,303,310,317,321

%N Positions of 1 in A287372; complement of A287527.

%C 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 From _Michel Dekking_, Mar 18 2018: (Start)

%C 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 sigma(1) = 2, sigma(2) = 3, sigma(3) = 2112,

%C and delta is the 'decoration' morphism defined by

%C delta(1) = 00, delta(2) = 1000, delta(3) = 0001000.

%C Let M be the incidence matrix of the morphism sigma, i.e., M equals

%C |0 0 2|

%C |1 0 2|

%C |0 1 0|.

%C 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 Solving chi(u) = 0, one finds that

%C LPF = (1/3)*(27+3*sqrt(57))^(1/3)+2/(27+3*sqrt(57))^(1/3).

%C For the frequencies one computes

%C mu[1] = 2/D, mu[2] = LPF^2/D, and mu[3] = LPF/D,

%C where D = LPF^2+LPF+2.

%C 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 To find the value of m, note that there are

%C A(n):= N(2)(sigma^n(1)) + N(3)(sigma^n(1))

%C 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 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 A(n)/|SR^n(00)| -> m as n->infinity,

%C where |SR^n(00)| denotes the length of SR^n(00).

%C But

%C |SR^n(00)| = 2N(1)(sigma^n(1)) + 4N(2)(sigma^n(1)) +7N(3)(sigma^n(1)).

%C It follows therefore that

%C m = (mu[1]+mu[3])/(2mu[1]+4mu[2]+7mu[3]) = 5.899687789...

%C (End)

%H Clark Kimberling, <a href="/A287402/b287402.txt">Table of n, a(n) for n = 1..10000</a>

%t s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];

%t w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "000"}]

%t Table[w[n], {n, 0, 8}]

%t st = ToCharacterCode[w[22]] - 48 (* A287372 *)

%t Flatten[Position[st, 0]] (* A287527 *)

%t Flatten[Position[st, 1]] (* A287402 *)

%Y Cf. A287372, A287527.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jun 17 2017