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A286687
Positions of 1 in A286685; complement of A286686.
4
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
1,1
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)
LINKS
EXAMPLE
As a word, A286685 = 01011001011001010110010110..., in which 0 is in positions 1,3,6,7,9,12,...
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 13 2017
STATUS
approved