A285951 Positions of 1's in A285949; complement of A285950.

2, 6, 9, 11, 15, 17, 20, 24, 27, 29, 32, 36, 38, 42, 45, 47, 51, 53, 56, 60, 62, 66, 69, 71, 74, 78, 81, 83, 87, 89, 92, 96, 99, 101, 104, 108, 110, 114, 117, 119, 122, 126, 129, 131, 135, 137, 140, 144, 146, 150, 153, 155, 159, 161, 164, 168, 171, 173, 176

Offset: 1

Author

Clark Kimberling, May 02 2017

Keywords

• nonn
• easy

Comments

Conjecture: 3n - a(n) is in {0, 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 1:

A285949 = 0|1000|100|10|1000|10|100| ....

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

The three return words are u:=10, v:=100 and w:=1000. These words uniquely correspond to the conjugated three words u'=01, v'=010, w'=0100 in A285949, which are the unique images u'=T(0), v'=T(01) and w'=T(011) of the words 0, 01 and 011 in the Thue-Morse word A010060. The images of these three words under the Thue-Morse morphism 0->01, 1->10 are 01, 0110 and 011010, and we have

T(01)=010, T(0110)=010001, T(011010)=010001001.

Shifting by 1 in A285949, these correspond uniquely to the conjugated words 100, 100010, and 100010010. It follows that the Thue-Morse morphism induces the morphism u->v, v->wu, w->wvu on the return words.

This morphism is modulo a change of alphabet equal to the ternary Thue-Morse morphism with fixed point A007413.

Note that on the alphabet {4,3,2} of the respective lengths of w, v, and u we obtain the sequence (a(n+1)-a(n)) = 4,3,2,4,2,3,4,3,2,... of first differences of the positions of the 1's in A285949.

To prove the formula a(n) = A010060(n)+ 3n-1, it suffices to show that a(n+1)-a(n) = A010060(n+1)-A010060(n)+3.

That this indeed is true: see the Comments of A029883, the first differences of the standard form of the Thue-Morse sequence A001285.

(End)

Examples

			As a word, A285949 = 0100010010100010100100010..., in which 1 is in positions  2,6,9,11,...
		

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, A285950, A285952.

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"}]  (* A284949, word *)
st = ToCharacterCode[w1] - 48 (* A284949, sequence *)
Flatten[Position[st, 0]] (* A285950 *)
Flatten[Position[st, 1]] (* A285951 *)

Python

def A285951(n): return ((n-1).bit_count()&1)+3*n-1 # Chai Wah Wu, May 21 2025

Formula

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