OFFSET
1,1
COMMENTS
Conjecture: 2 < n*r - a(n) < 3 for n >= 1, where r = (7 + sqrt(5))/2.
From Michel Dekking, Feb 12 2021: (Start)
Let T be the transform given by T(0) = 1, T(1) = 011 that defines A287725.
The Fibonacci word A003849 is a 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(010) = 10111, T(sigma^2(1)) = T(01) = 1011.
We see from this that the sequence (a(n+1)-a(n)) of first differences 5,4,5,5,4,5,4,5,5,4,5,5,4..., of (a(n)) is a sequence on the two 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(n)) is the generalized Beatty sequence given by a(n) = floor(n*phi) + 3n - 2.
This immediately implies Kimberling's conjecture.
Note that r = (7 + sqrt(5))/2 = phi + 3, where phi is the golden mean. So
n*r - a(n) = n(phi+3) - floor(n*phi) + 3n - 2 = n*phi - floor(n*phi) - 2,
which lies in (2,3).
These bounds are best possible 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(n) = floor(n*phi) + 3n - 2. - Michel Dekking, Feb 12 2021
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 02 2017
STATUS
approved