OFFSET
1,1
COMMENTS
Conjecture: a(n)/n-> 3 + sqrt(5), and if m denotes this number, then -1 < m - a(n)/n) < 3 for n >= 1.
From Michel Dekking, Oct 19 2018: (Start)
Here is a proof of this conjecture. Note that
q(n) := n/a(n) = [a(1)+a(2)+...+a(n)]/a(n)
is the frequency of 1's in the first a(n) terms of the sequence.
It follows from the sequel that (q(n)) converges as n->infinity.
So we have to show that
q(n) --> 1/m = (3-sqrt(5))/4.
It is useful here to profit from Kimberling's observation in the Comments of x:=A288707 that x is the {0->00, 1->10} transform of the morphism 0->10, 1->0.
We see from this that a 1 occurs at position 2k-1 in x if
and only if a 1 occurs at position k in the fixed point
y = A189661 = 0, 1, 0, 1, 0, 0, 1, 0, 1, 0,...
with y(1)=0 of the square of the time reversed Fibonacci morphism 0->10,1->0 (this explains why all the numbers in (a(n)) are odd).
Using the incidence matrix of the morphism 0->010, 1->10,
one can calculate that the frequency of 1's in y equals (3-sqrt(5))/2,
and so the frequency of 1's in x is half this number, and we have proved that
q(n) --> (3-sqrt(5))/4.
The bounds -1 < m - a(n)/n < 3 are equivalent to bounds on
2q(n)-(3-sqrt(5))/2.
The latter can be proved by checking them for a finite number of n, and then using the exponential convergence of the 2q(n) to (3-sqrt(5))/2 (a consequence of the Perron-Frobenius theorem). (End)
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 16 2017
STATUS
approved