



1, 2, 5, 6, 7, 10, 11, 14, 15, 16, 19, 20, 21, 24, 25, 28, 29, 30, 33, 34, 37, 38, 39, 42, 43, 44, 47, 48, 51, 52, 53, 56, 57, 58, 61, 62, 65, 66, 67, 70, 71, 74, 75, 76, 79, 80, 81, 84, 85, 88, 89, 90, 93, 94, 97, 98, 99, 102, 103, 104, 107, 108, 111, 112
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OFFSET

1,2


COMMENTS

a(n)  a(n1) is in {1,2,3} for n>=2, and a(n)/n > 4  sqrt(5).
From Michel Dekking, Aug 29 2020: (Start)
This sequence is a generalized Beatty sequence.
Recall from A286749 that A286749 is the lettertoletter image of the fixed point x of the morphism mu given by
mu: 1>12341, 2>1, 3>2, 4>34.
where the lettertoletter map lambda is defined by
lambda: 1>1, 2>1, 3>0, 4>0.
The return words of the word 1 in x are A:=1 and B:=1234.
We have mu(1)=12341, and mu(1234)=123411234. So the derived morphism is tau: A>BA, B>BAB.
This morphism happens to be the square of the Fibonacci morphism on the alphabet {B,A}.
The return word A = 1 has lambdaimage 1, and the return word B = 1234 has lambdaimage 1100. This means that they give distances 1, respectively 1 and 3 between (successive) occurrences of 1's in A286749. This leads to the decoration B>13, A>1, which amounts to replacing 0 by 01 and 1 by 0 in the Fibonacci word.
But the Fibonacci word is fixed by the {0,1} version of tau. It follows that the sequence (a(n+1)a(n)) = 1,3,1,1,3,... is the Fibonacci word on the alphabet {1,3}. Finally, Lemma 8 in the paper "Generalized Beatty sequences..." then gives that a(n) = 5*n2*floor(n*phi)2, for n>0.
Clearly this implies a(n)/n > 4  sqrt(5).
(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], 20182019.


FORMULA

a(n) = 5*n2*floor(n*phi)2.  Michel Dekking, Aug 29 2020


EXAMPLE

As a word, A286749 = 11001110011001110011010..., in which 1 is in positions 1,2,5,6,7,10,...


MATHEMATICA

s = Nest[Flatten[# /. {0 > {0, 1}, 1 > {0}}] &, {0}, 12]; (* A003849 *)
w = StringJoin[Map[ToString, s]];
w1 = StringReplace[w, {"0100" > ""}]; st = ToCharacterCode[w1]  48; (* A286749 *)
Flatten[Position[st, 0]]; (* A286750 *)
Flatten[Position[st, 1]]; (* A286751 *)


CROSSREFS

Cf. A003849, A286749, A286750.
Sequence in context: A039015 A037453 A014528 * A293278 A087791 A334880
Adjacent sequences: A286748 A286749 A286750 * A286752 A286753 A286754


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, May 14 2017


STATUS

approved



