



2, 3, 7, 8, 11, 12, 16, 17, 21, 22, 25, 26, 30, 31, 34, 35, 39, 40, 44, 45, 48, 49, 53, 54, 58, 59, 62, 63, 67, 68, 71, 72, 76, 77, 81, 82, 85, 86, 90, 91, 94, 95, 99, 100, 104, 105, 108, 109, 113, 114, 118, 119, 122, 123, 127, 128, 131, 132, 136, 137, 141
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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(2n1)) 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(2n1)) is the generalized Beatty sequence given for n = 1,2,... by
a(2n1) = 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
(2n1)*r  a(2n1) = (n1/2)(phi+3)  floor(n*phi)  3n + 2 =
n*phi  floor(n*phi)  1/2*phi + 1/2,
which lies in the interval (1/2phi/2, 3/2phi/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 (1/2phi/2, 3/2phi/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, 325341, 2019.


FORMULA

a(2n1) = floor(n*phi) + 3n  2 , a(2n) = floor(n*phi) + 3n  1 .  Michel Dekking, Feb 12 2021
a(2n1) = A287726(n), a(2n) = A287726(n) + 1.  Michel Dekking, Feb 12 2021


MATHEMATICA

s = Nest[Flatten[# /. {0 > {0, 1}, 1 > {0}}] &, {0}, 10] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" > "1", "1" > "001"}]
st = ToCharacterCode[w1]  48 (* A287674 *)
Flatten[Position[st, 0]] (* A287675 *)
Flatten[Position[st, 1]] (* A287676 *)


CROSSREFS

Cf. A287675, A287676, A287726.
Sequence in context: A051468 A002274 A329783 * A273944 A214324 A102664
Adjacent sequences: A287672 A287673 A287674 * A287676 A287677 A287678


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Jun 02 2017


STATUS

approved



