OFFSET
1,1
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
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 17 2017
STATUS
approved