OFFSET

1,2

COMMENTS

Conjecture: a(n)/n -> 4/3.

From Michel Dekking, Apr 19 2022: (Start)

Kimberling's conjecture is equivalent to the property that the frequency of 1's in A285960 is equal to 3/4. This follows directly from the corresponding result in A285961.

But more is true. The first difference sequence (d(n)) = 1,1,2,1,2,1,1,1,2,1,1,2,1,... of (a(n)) is a morphic sequence.

From the representation of A285960 by the decoration A-> 11, B-> 10, C-> 110, D-> 1, we see that the differences between occurrences of 1's are also given by a decoration:

A->11, B->2, C->12, D->1.

This time one finds that (d(n)) can be obtained as a letter to letter image of a morphic sequence fixed point of a morphism mu on {a,b,c,d,e} given by

mu: a->ab, b->c, c->ade, d->ce, e->ad,

with the letter-to-letter map

lambda: a->1, b->1, c->2, d->2, e->1.

We have (d(n)) = lambda(z), where z is the fixed point z = abcade... of mu.

(End)

LINKS

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

EXAMPLE

As a word, A285960 = 111011011110111011..., in which 1 is in positions 1,2,3,5,6,8,...

MATHEMATICA

CROSSREFS

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 05 2017

STATUS

approved