A298952 First put a(n)=0 for all n, then start with a(0) = 1 and add at step n >= 0 the term 1 at position 2*n + a(n).

1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0

Offset

0

Comments

Sum_{i = 0..n} a(i)/n tends to 1/2 as n tends to infinity. [corrected by Rémy Sigrist, Jan 31 2018]

From Michel Dekking, Sep 07 2020: (Start)

The above limit statement follows from a much stronger property.

Let mu be the 'exchanged' Thue-Morse morphism given by

mu(0) = 10, mu(1) = 01.

CLAIM: a(234...) = mu(a(123...)).

Here a(234...) denotes the word associated to the sequence a(2), a(3), a(4),....

Proof: If a(n)=1 then a(2n+1)=1, and it also follows that a(2n)=0. If a(n)=0 then a(2n)=1, and it also follows that a(2n+1)=0.

This can also be expressed as mu(a(n))=a(2n)a(2n+1).

(End)

Links

Table of n, a(n) for n=0..120.

Example

Set a(n) = 0.
n = 0, a(0) = 1. Add term 1 at position 2*0+1 = 1. We have {1,1,0,0,0,0,0,0,0,0,...}
n = 1, a(1) = 1. Add term 1 at position 2*1+1 = 3. We have {1,1,0,1,0,0,0,0,0,0,...}
n = 2, a(2) = 0. Add term 1 at position 2*2+0 = 4. We have {1,1,0,1,1,0,0,0,0,0,...}
n = 3, a(3) = 1. Add term 1 at position 2*3+1 = 7. We have {1,1,0,1,1,0,0,1,0,0,...}
and so on.

Prog

(PARI) a(n) = if(n==0, 1, (logint(n, 2) - hammingweight(n)) % 2); \\ Kevin Ryde, Mar 11 2021

Crossrefs

Cf. A059448 (complement), A242179 (values +-1).

Indices of 0's and 1's (except n=0): A059009, A059010.

Cf. A298307.

Sequence in context: A167753 A353816 A141727 * A123594 A286801 A189091

Adjacent sequences: A298949 A298950 A298951 * A298953 A298954 A298955

Keyword

nonn

Author

Ctibor O. Zizka, Jan 30 2018

Status

approved