



1, 2, 3, 5, 6, 8, 9, 10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 24, 25, 26, 28, 29, 30, 32, 33, 34, 35, 37, 38, 40, 41, 42, 43, 45, 46, 47, 49, 50, 51, 53, 54, 55, 56, 58, 59, 61, 62, 63, 65, 66, 67, 69, 70, 72, 73, 74, 75, 77, 78, 79, 81, 82, 83, 85, 86, 88
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OFFSET

1,2


COMMENTS

Conjecture: a(n)/n > 4/3.


REFERENCES

From Michel Dekking, Apr 19 2022: (Start)
Kimberling's conjecture is equivalent to the property that the frequency of 1's in A285960 is equal to 3/4. This follows directly from the corresponding result in A285961.
But more is true. The first difference sequence (d(n)) = 1,1,2,1,2,1,1,1,2,1,1,2,1,... of (a(n)) is a morphic sequence.
From the representation of A285960 by the decoration A> 11, B> 10, C> 110, D> 1, we see that the differences between occurrences of 1's are also given by a decoration:
A>11, B>2, C>12, D>1.
This time one finds that (d(n)) can be obtained as a letter to letter image of a morphic sequence fixed point of a morphism mu on {a,b,c,d,e} given by
mu: a>ab, b>c, c>ade, d>ce, e>ad,
with the lettertoletter map
lambda: a>1, b>1, c>2, d>2, e>1.
We have (d(n)) = lambda(z), where z is the fixed point z = abcade... of mu.
(End)


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000


EXAMPLE

As a word, A285960 = 111011011110111011..., in which 1 is in positions 1,2,3,5,6,8,...


MATHEMATICA

s = Nest[Flatten[# /. {0 > {0, 1}, 1 > {1, 0}}] &, {0}, 8] (* ThueMorse, A010060 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"01" > "1"}]
st = ToCharacterCode[w1]  48 (* A285960 *)
Flatten[Position[st, 0]] (* A285961 *)
Flatten[Position[st, 1]] (* A285962 *)


CROSSREFS

Cf. A010060, A285960, A285961.
Sequence in context: A039104 A184861 A183572 * A047450 A039073 A026359
Adjacent sequences: A285959 A285960 A285961 * A285963 A285964 A285965


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, May 05 2017


STATUS

approved



