OFFSET
1,1
COMMENTS
Conjecture: -2 < n*r - a(n) < 1 for n >= 1, where r = (7 + sqrt(5))/6. [Corrected by Clark Kimberling, Aug 19 2019]
From Michel Dekking, Aug 22 2019: (Start)
Computation of a formula for a(n): we will show that (a(n)) is a union of three generalized Beatty sequences.
Let T denote the morphism {0->1, 1->000}.
The Fibonacci word xF:=A003849 is a concatenation of the two words v = 100 and w = 10 (ignoring xF(0)). It is known that the order of these words is given by the Fibonacci word itself (cf. A003622). Now note that
T(v) = T(100) = 00011, T(w) = T(10) = 0001.
We see from this that the sequence of first differences of a = A287664 for the first 0 in each triple 000 is given by 5,4,5,5,4,5,4,5,5,4,..., and that in fact these two letters occur as the Fibonacci word on the alphabet {5,4}.
Let {b^(j)} for j = 1,2,3 be the sequence of positions of the j-th 0 in the triples 000. From Lemma 8 in the paper by Allouche and Dekking it follows that these sequences are given by
b^(j)(n) = floor(n*phi) + 3*n + r^(j),
where r^(1) = -2, r^(2) = -1, and r^(3) = 0.
Note that a(3*n+j-3) = b^(j)(n) for n=1,2,3,... and j=1,2,3. (End)
From Michel Dekking, Aug 22 2019: (Start)
Proof of Kimberling's conjecture. {} denotes fractional part.
Let r = (7+sqrt(5))/6 = phi/3+1. Using the formula above we have
(3*n+j-3)*r - a(3*n+j-3) =
(n+j/3-1)*(phi+3) - (n*phi - {n*phi} + 3n + r^(j)) =
(j/3-1)*phi + j - 3 - r^(j) + {n*phi}.
Since ({n*phi}) is equidistributed on (0,1), this implies that a(3*n+j-3) takes values in the interval (-(2/3)*phi, -(2/3)*phi+1) for j=1, in the interval (-(1/3)*phi, -(1/3)*phi+1) for j=2, and in the interval (0,1) for j=3. Combining these, we see that a(n) takes values in (-(2/3)*phi, 1), and that this is best possible. Since -(2/3)*phi = -1.078689..., this improves on the lower bound conjectured by Kimberling.
(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(3*n+j-3) = b^(j)(n) for j=1,2,3, with b^(1)(n) = floor(n*phi) + 3*n - 2, b^(2)(n) = floor(n*phi) + 3*n - 1, b^(3)(n) = floor(n*phi) + 3*n. - Michel Dekking, Aug 22 2019
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 02 2017
STATUS
approved