



3, 7, 13, 17, 23, 29, 33, 39, 43, 49, 55, 59, 65, 71, 75, 81, 85, 91, 97, 101, 107, 111, 117, 123, 127, 133, 139, 143, 149, 153, 159, 165, 169, 175, 181, 185, 191, 195, 201, 207, 211, 217, 221, 227, 233, 237, 243, 249, 253, 259, 263, 269, 275, 279, 285, 289
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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 = (3sqrt(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 2k1 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 (3sqrt(5))/2,
and so the frequency of 1's in x is half this number, and we have proved that
q(n) > (3sqrt(5))/4.
The bounds 1 < m  a(n)/n < 3 are equivalent to bounds on
2q(n)(3sqrt(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 (3sqrt(5))/2 (a consequence of the PerronFrobenius theorem). (End)


LINKS

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


MATHEMATICA

s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n  1], {"00" > "1000", "10" > "00"}]
Table[w[n], {n, 0, 8}]
st = ToCharacterCode[w[10]]  48 (* A288707 *)
Flatten[Position[st, 0]] (* A288708 *)
Flatten[Position[st, 1]] (* A288709 *)


CROSSREFS

Cf. A288707, A288708.
Sequence in context: A285669 A063239 A063226 * A040106 A191028 A034914
Adjacent sequences: A288706 A288707 A288708 * A288710 A288711 A288712


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Jun 16 2017


STATUS

approved



