A287402 Positions of 1 in A287372; complement of A287527.

4, 8, 12, 19, 23, 31, 35, 43, 50, 54, 58, 65, 69, 77, 84, 91, 95, 103, 107, 115, 122, 129, 133, 141, 148, 152, 156, 163, 167, 175, 179, 187, 194, 198, 202, 209, 213, 221, 228, 235, 239, 247, 254, 258, 262, 269, 276, 280, 284, 291, 295, 303, 310, 317, 321

Offset: 1

Author

Clark Kimberling, Jun 17 2017

Keywords

• nonn
• easy

Comments

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.

From Michel Dekking, Mar 18 2018: (Start)

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

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

and delta is the 'decoration' morphism defined by

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

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

|0 0 2|

|1 0 2|

|0 1 0|.

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.

Solving chi(u) = 0, one finds that

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

For the frequencies one computes

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

where D = LPF^2+LPF+2.

From the existence of these frequencies one can deduce the existence of the limit m of a(n)/n as n tends to infinity.

To find the value of m, note that there are

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

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.

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

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

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

But

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

It follows therefore that

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

(End)

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A287372, A287527.

Programs

Mathematica

s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "000"}]
Table[w[n], {n, 0, 8}]
st = ToCharacterCode[w[22]] - 48   (* A287372 *)
Flatten[Position[st, 0]]  (* A287527 *)
Flatten[Position[st, 1]]  (* A287402 *)