OFFSET
1,1
COMMENTS
Conjecture: -1 < n*r - a(n) < 2 for n >= 1, where r = (7 + sqrt(5))/4.
From Michel Dekking, Feb 12 2021: (Start)
Let T be the transform given by T(0) = 1, T(1) = 001 that defines the underlying sequence A287674 by A287674(n) = T(A003849(n)).
The Fibonacci word A003849 is fixed point of the morphism sigma: 0->01, 1->0, and therefore also of the morphism sigma^2: 0->010, 1->01.
Now note that
T(sigma^2(0)) = T(10011) = 10111, T(sigma^2(1)) = T(01) = 1001.
We see from this that the sequence (a(2n+1)-a(2n-1)) of first differences 5,4,5,5,4,4,5,4,5,5,4,..., of every second occurrence of a 0 is a sequence on the letters 4 and 5, and that in fact these two letters occur as the Fibonacci word on the alphabet {5,4}.
From Lemma 8 in the paper by Allouche and Dekking it follows that (a(2n-1)) is the generalized Beatty sequence given for n = 1,2,... by
a(2n-1) = floor((n*phi) + 3n - 2 = A287726(n),
and (a(2n)) is the generalized Beatty sequence given for n = 1,2,... by
a(2n) = floor(n*phi) + 3n - 1.
From this we derive directly Kimberling's conjecture. Note that
r = (7 + sqrt(5))/4 = phi/2 + 3/2,
where phi is the golden mean. So
(2n-1)*r - a(2n-1) = (n-1/2)*(phi+3) - floor(n*phi) - 3n + 2 =
n*phi - floor(n*phi) - (1/2)*phi + 1/2,
which lies in the interval (1/2 - phi/2, 3/2 - phi/2). Also we have
2n*r - a(2n) = n*(phi+3) - floor(n*phi) - 3n + 1 =
n*phi - floor(n*phi) + 1,
which lies in the interval (1,2).
The two intervals (1/2 - phi/2, 3/2 - phi/2) = (-0.3090..., 0.6090...) and (1,2) are subintervals of the interval (-1,2) in Kimberling's conjecture, and there are no better bounds, since the fractional part of n*phi is equidistributed modulo 1.
(End)
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
J.-P. Allouche and F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424 [math.NT], 2018.
J.-P. Allouche and F. M. Dekking, Generalized Beatty sequences and complementary triples, Moscow J. Comb. Number Th. 8, 325-341, 2019.
FORMULA
a(2n-1) = floor(n*phi) + 3n - 2, a(2n) = floor(n*phi) + 3n - 1. - Michel Dekking, Feb 12 2021
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 02 2017
STATUS
approved