OFFSET
1,2
COMMENTS
Conjecture: 0 < n*r - a(n) < 3 for n >= 1, where r = (-1 + 3*sqrt(5))/2. [Corrected by Clark Kimberling, Aug 19 2019]
From Michel Dekking, Aug 21 2019: (Start)
Proof of Maiga's conjecture: let T denote the morphism {0->1, 1->000}.
The Fibonacci word xF:=A003849 is fixed point of the morphism 0->01, 1->0 and therefore xF is a concatenation of the two words v=01 and w=0, where these words occur as the Fibonacci word itself.
Now note that
T(v) = T(01) = 1000, T(w) = T(0) = 1.
We see from this that the sequence of first differences of A287665, Delta A287665 = 4,1,4,4,1,4,1,4,4,1,..., is a sequence on the letters 4 and 1, and that in fact these two letters occur as the Fibonacci word on the alphabet {4,1}.
Since A001468 (starting from n=1) is the Fibonacci word on the alphabet {2,1}, Maiga's formula follows.
Proof of Kimberling's conjecture.
It follows from the result above by Lemma 8 in the Allouche-Dekking paper that A287665 is a generalized Beatty sequence
a(n) = 3*floor(n*phi) - 2*n.
So if r = (-1 + 3*sqrt(5))/2 = 3*phi - 2, then
n*r - a(n) = n*(3*phi-2) - [3*(n*phi-{n*phi})]-2*n = 3*{n*phi}, where {} denotes fractional part.
It follows that 0 < n*r - a(n) < 3. Moreover, these bounds are tight, since the sequence ({n*phi}) is equidistributed on (0,1).
(End)
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
J.-P. Allouche, F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424 [math.NT], 2018.
FORMULA
a(n) = 3*floor(n*phi) - 2*n. - Michel Dekking, Aug 21 2019
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 02 2017
STATUS
approved