A287770 Positions of 1 in A287769; complement of A276855.

1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 14, 15, 16, 17, 19, 20, 21, 22, 24, 25, 26, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 47, 48, 49, 51, 52, 53, 54, 56, 57, 58, 59, 61, 62, 63, 65, 66, 67, 68, 70, 71, 72, 74, 75, 76, 77, 79, 80, 81, 82, 84

Offset

1,2

Comments

Conjecture: 0 < n*r - a(n) < 1 for n >= 1, where r = (15 - sqrt(5))/10. It has been verified by computer that a(n) = floor(n*r) for n=1..3*10^6.

This conjecture can be proved from the result in the Comments of A287769, where it is shown that A287769 is a Sturmian sequence with slope s := 1 - 1/(3+phi) = (15+sqrt(5))/22. The conjecture then follows from Lemma 9.1.3 in "Automatic sequences", since r = 1/s. - Michel Dekking, Oct 11 2017

References

Jean-Paul Allouche and Jeffrey Shallit, Automatic sequences, Theory, applications, generalizations, Cambridge University Press, Cambridge, 2003, xvi+571.

Links

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

Formula

a(n) = floor(n*r), where r = (15 - sqrt(5))/10. - Michel Dekking, Oct 11 2017

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "1", "1" -> "110"}]
st = ToCharacterCode[w1] - 48    (* A287769 *)
Flatten[Position[st, 0]]  (* A276855 *)
Flatten[Position[st, 1]]  (* A287770 *)

Crossrefs

Cf. A276855, A287769.

Sequence in context: A284955 A093480 A121518 * A039080 A093435 A048964

Adjacent sequences: A287767 A287768 A287769 * A287771 A287772 A287773

Keyword

nonn,easy

Author

Clark Kimberling, Jun 03 2017

Status

approved