OFFSET
1,1
COMMENTS
Conjecture: 2n - a(n) is in {0,1} for n >= 1.
From Michel Dekking, Apr 12 2022: (Start)
Obviously Kimberling's conjecture is equivalent to the property that A286046 is a concatenation of the two 2-blocks 01 and 10. This can be read off immediately from the {A, B, C, D} composition of A286046 given in the comments of that sequence.
But more is true. The first difference sequence (d(n)) = 2,2,1,3,2,1,3,... of (a(n)) is a morphic sequence. From the representation of A286046 by the decoration A->1010, B->1001, C->101001, D->10, we see that the differences between occurrences of 0's are given by a decoration:
A->22, B->13, C->213, D->2.
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 an 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->2, b->2, c->1, d->3, e->1, f->2.
We have (d(n)) = lambda(z), where z is the fixed point z = abcdae... of mu.
(End)
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
EXAMPLE
As a word, A286046 = 101010011010011010..., in which 0 is in positions 2,4,6,7,10,...
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 07 2017
STATUS
approved