OFFSET
1,1
COMMENTS
Conjecture: lim_{n->oo} a(n)/n = 4.
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 1/4. This can be computed via the comments of A285960.
But more is true. The first difference sequence (d(n)) = 3,5,4,4,3,4,4,5,3,... of (a(n)) is a morphic sequence. Let
A=011, B=010, C=0110, and D=01.
From the representation of A285960 as fixed point of the derived morphism
tau: A -> AB, B -> CD, C -> ABD, D -> C,
decorated by the morphism
delta: A-> 11, B-> 10, C-> 110, D-> 1,
we do not see how to obtain the differences between occurrences of 0's. However, A285960 is also represented as fixed point of
tau^2: A->ABCD, B->ABDC, C->ABCDC, D->ABD, and delta, where the four images turn into
delta(tau^2(A))=11101101, delta(tau^2(B))=11101110,
delta(tau^2(C))=11101101110, delta(tau^2(D))=11101.
The corresponding consecutive distances between 0's are 35, 44, 344, and 5.
The "natural" algorithm to obtain (d(n)) as a letter-to-letter image of a morphic sequence from this decoration yields (for example) a morphism mu on a six-letter alphabet {a,b,c,d,e,f} given by
mu: a->ab, b->cd, c->aed, d->f, e->cd, f->aed,
with the letter-to-letter map
lambda: a->3, b->5, c->4, d->4, e->4, f->5.
We have (d(n)) = lambda(z), where z is the fixed point z = abcdaed... of mu.
Note that it turns out that mu is the same morphism as the one used to obtain A286047 and A286048 as morphic sequences.
Another way to prove Kimberling's conjecture: a(n)/n = (1/n)*Sum_{k=1..n} d(k), the average value M of the differences.
To obtain the limit, you need the frequencies M(a),...,M(f) of a,b,...,f in the fixed point of mu. These are computed with the Perron-Frobenius theorem.
We have M(a)=M(d)=1/4, M(b)=M(c)=M(d)=M(f)=1/8. So M = (2/8)*3 + (4/8)*4 + (2/8)*5 = (1/8)*(6+16+10) = 32/8 = 4.
(End)
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
EXAMPLE
As a word, A285960 = 111011011110111011..., in which 0 is in positions 4,7,12,16,...
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 05 2017
STATUS
approved