A286686 Positions of 0 in A286685; complement of A286687.

1, 3, 6, 7, 9, 12, 13, 15, 17, 20, 21, 23, 26, 27, 29, 31, 34, 35, 37, 40, 41, 43, 46, 47, 49, 51, 54, 55, 57, 60, 61, 63, 65, 68, 69, 71, 74, 75, 77, 80, 81, 83, 85, 88, 89, 91, 94, 95, 97, 99, 102, 103, 105, 108, 109, 111, 113, 116, 117, 119, 122, 123, 125

Offset: 1

Author

Clark Kimberling, May 13 2017

Keywords

• nonn
• easy

Comments

a(n) - a(n-1) is in {1,2,3} for n>=2, and a(n)/n -> 2. These are also the positions of 1 in the {0->10, 1->01}-transform of the Pell word, A171588.

From Michel Dekking, Sep 19 2019: (Start)

Here is a precise description of the sequence of first differences.

Let tau be the map tau: 0->01, 1->10. By definition A286685 equals tau(b), where b is the Pell word. The words of length 2 occurring in b are 00, 01 and 10. These are mapped by tau to

tau(00) = 0101, tau(01) = 0110, tau(10) = 1001.

Each of these three four letter words contains exactly 2 0's, occurring among the first two letters and among the last two letters. It follows from this that the overlapping words of length 2 in the Pell word b induce distances between 0's in tau(b) of 2 for 00, of 3 for 01, and of 1 for 10. But then the difference sequence (a(n+1) - a(n)) = 2, 3, 1, 2, 3, 1, 2, ... is equal to the 1->3, 2->1, 3->2 permuted version of the 3-symbol Pell word A294180. (End)

Examples

			As a word, A286685 = 01011001011001010110010110..., in which 0 is in positions 1,3,6,7,9,12,...
		

Links

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

Crossrefs

Cf. A171588, A286685, A286687.

Programs

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 0, 1}, 1 -> {0}}] &, {0}, 6] (* A171588 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "01", "1"->"10"}]
st = ToCharacterCode[w1] - 48 ; (* A286685 *)
p0 = Flatten[Position[st, 0]];  (* A286686 *)
p1 = Flatten[Position[st, 1]];  (* A286687 *)