A285954 Positions of 1 in A285952; complement of A285953.

1, 2, 4, 6, 7, 9, 10, 11, 13, 15, 16, 17, 19, 20, 22, 24, 25, 27, 28, 29, 31, 32, 34, 36, 37, 38, 40, 42, 43, 45, 46, 47, 49, 51, 52, 53, 55, 56, 58, 60, 61, 62, 64, 66, 67, 69, 70, 71, 73, 74, 76, 78, 79, 81, 82, 83, 85, 87, 88, 89, 91, 92, 94, 96, 97, 99

Offset

1,2

Comments

Conjecture: 3n/2 - a(n) is in {0, 1/2, 1} for n >= 1.

Proof of the conjecture: Let t=A010060 be the Thue-Morse sequence. Every pair t(2n-1),t(2n) is either 01 or 10. Since 01 and 10 map to 110 and 101 under the transform, which both have length 3, it follows that a(2n+1) = 3n+1, and a(2n) = 3n if t(2n)=0, a(2n) = 3n-1 if t(2n)=1 for n=1,2,..., and so certainly 3n/2 - a(n) is 0, 1/2 or 1. - Michel Dekking, Jan 05 2018

Links

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

Formula

a(2n+1) = 3n+1, a(2n) = 3n - A010060(2n) - Michel Dekking, Jan 05 2018

a(n) = n+floor(n/2) if n is odd and a(n) = n+floor(n/2)-A010060(n-1) otherwise. - Chai Wah Wu, Nov 17 2024

Example

As a word, A285952 = 110101101110101..., in which 1 is in positions 1,2,4,6,7,9,...

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7]  (* Thue-Morse, A010060 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "1", "1" -> "10"}]  (* A285952, word *)
st = ToCharacterCode[w1] - 48 (* A285952, sequence *)
Flatten[Position[st, 0]]  (* A285953 *)
Flatten[Position[st, 1]]  (* A285954 *)

Prog

(Python)
def A285954(n): return n+(n>>1)-(0 if n&1 else (n-1).bit_count()&1) # Chai Wah Wu, Nov 17 2024

Crossrefs

Cf. A010060, A005614, A285950, A285952, A285953.

Sequence in context: A186493 A289072 A287379 * A342778 A289075 A111094

Adjacent sequences: A285951 A285952 A285953 * A285955 A285956 A285957

Keyword

nonn,easy

Author

Clark Kimberling, May 05 2017

Status

approved