A285950 Positions of 0's in A285949; complement of A285951.

1, 3, 4, 5, 7, 8, 10, 12, 13, 14, 16, 18, 19, 21, 22, 23, 25, 26, 28, 30, 31, 33, 34, 35, 37, 39, 40, 41, 43, 44, 46, 48, 49, 50, 52, 54, 55, 57, 58, 59, 61, 63, 64, 65, 67, 68, 70, 72, 73, 75, 76, 77, 79, 80, 82, 84, 85, 86, 88, 90, 91, 93, 94, 95, 97, 98

Offset: 1

Author

Clark Kimberling, May 02 2017

Keywords

• nonn
• easy

Comments

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

From Michel Dekking, Sep 03 2019: (Start)

Proof of the conjecture by Kimberling: more is true. Here follows a proof of the formula below. Let T be the transform T(01)=0, T(1)=0.

Consider the return word structure of A285949 for the word 0:

A285949 = 01| 0| 0| 01| 0| 01| 01| 0| 0| 01| 01| ....

[See Justin & Vuillon (2000) for definition of return word. - N. J. A. Sloane, Sep 23 2019]

The two return words are v:=0 and w:=01. Always v = T(1)and w = T(01) in this decomposition of the image T(A010060) of A010060 under the transform. It follows that the return words occur as the Thue-Morse word 21121221211... on the alphabet {2,1}. But the lengths of the return words corresponds to the differences between the indices where the 0's occur in A285949, which generate (a(n)).

As the Thue-Morse word is a concatenation of 12 and 21 which, considered as integers, both add to 3, it follows that a(2n+1) = 3n+1. Similarly, it follows that a(2n) = 3n - A010060(n-1).

(End)

Examples

			As a word, A285949 = 0100010010100010100100010..., in which 0 is in positions 1,3,4,5,7,...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000
• Jacques Justin and Laurent Vuillon, Return words in Sturmian and episturmian words, RAIRO-Theoretical Informatics and Applications 34.5 (2000): 343-356.

Crossrefs

Cf. A010060, A003849, A285949, A285952.

Cf. A285951 (complement), A076826 (mod 3).

Programs

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7]  (* Thue-Morse, A010060 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "01", "1" -> "0"}]  (* A285949, word *)
st = ToCharacterCode[w1] - 48 (* A285949, sequence *)
Flatten[Position[st, 0]]  (* A285950 *)
Flatten[Position[st, 1]]  (* A285951 *)

PARI

a(n) = (3*n)>>1 - if(n%2==0, hammingweight(n-2)%2); \\ Kevin Ryde, Jan 28 2022

Python

def A285950(n): return 3*(n>>1)+(1 if n&1 else -((n-1>>1).bit_count()&1)) # Chai Wah Wu, May 21 2025

Formula

a(2n) = 3n - A010060(n-1); a(2n+1) = 3n + 1. - Michel Dekking, Sep 03 2019