A179854 Number of 0's (mod 3) in the binary expansion of n.

0, 1, 0, 2, 1, 1, 0, 0, 2, 2, 1, 2, 1, 1, 0, 1, 0, 0, 2, 0, 2, 2, 1, 0, 2, 2, 1, 2, 1, 1, 0, 2, 1, 1, 0, 1, 0, 0, 2, 1, 0, 0, 2, 0, 2, 2, 1, 1, 0, 0, 2, 0, 2, 2, 1, 0, 2, 2, 1, 2, 1, 1, 0, 0, 2, 2, 1, 2, 1, 1, 0, 2, 1, 1, 0, 1, 0, 0, 2, 2, 1, 1, 0, 1, 0, 0, 2, 1, 0, 0, 2, 0, 2, 2, 1, 2, 1, 1, 0, 1, 0, 0, 2, 1, 0, 0, 2, 0, 2, 2, 1, 1, 0, 0, 2, 0, 2, 2, 1, 0

Offset

1,4

Comments

A ternary analog of A059448.

Offset is 1 to avoid the ambiguity at n=0.

Inspired by Chapter 1 of Allouche and Shallit.

From Michel Dekking, Sep 30 2020: (Start)

Let tau be the "twisted" 3-symbol length 2 Thue-Morse morphism given by

tau(0) = 10, tau(1) = 21, tau (2) = 02.

The name of tau is in analogy with the comments from A297531. The "ordinary" 3-symbol length 2 Thue-Morse morphism is the morphism mu given by

mu(0) = 01, mu(1) = 12, mu(2) = 20.

The unique fixed point of mu is the sequence A071858 = 01121220...

We have mu^3 = tau^3.

The sequence a = (a(n)) satisfies

a = 0 tau(a).

This follows directly from the recursion formulas

a(2n) = a(n) + 1 mod 3, a(2n+1) = a(n).

(End)

References

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.

Links

Table of n, a(n) for n=1..120.

Formula

a(2n) = a(n) + 1 mod 3, a(2n+1) = a(n). - Michel Dekking, Sep 30 2020

Maple

s1:=[];
for n from 0 to 200 do
t1:=convert(n, base, 2); t2:=subs(1=NULL, t1); s1:=[op(s1), nops(t2) mod 3]; od:
s1;

Crossrefs

Cf. A059448. Related to A071858.

Sequence in context: A157044 A182748 A025879 * A250622 A212551 A125753

Adjacent sequences: A179851 A179852 A179853 * A179855 A179856 A179857

Keyword

nonn

Author

N. J. A. Sloane, Jan 11 2011

Status

approved