A286687 Positions of 1 in A286685; complement of A286686.

2, 4, 5, 8, 10, 11, 14, 16, 18, 19, 22, 24, 25, 28, 30, 32, 33, 36, 38, 39, 42, 44, 45, 48, 50, 52, 53, 56, 58, 59, 62, 64, 66, 67, 70, 72, 73, 76, 78, 79, 82, 84, 86, 87, 90, 92, 93, 96, 98, 100, 101, 104, 106, 107, 110, 112, 114, 115, 118, 120, 121, 124

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 0 in the {0->10, 1->01}-transform of the Pell word, A171588.

From Michel Dekking, Feb 21 2018: (Start)

Let tau be transform 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 1'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 1's in tau(b) of 2 for 00, of 1 for 01, and of 3 for 10. But then the difference sequence (a(n+1) - a(n)) = 2, 1, 3, 2, 1, 3, 2, 2, 1, 3, 2, 1, 3, 2, 2, ... is equal to the 1->2, 2->1 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, A286686.

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 *)
Flatten[Position[st, 0]];  (* A286686 *)
Flatten[Position[st, 1]];  (* this sequence *)