A395958 Third iterate of the Thue-Morse transform applied to A000035.

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

Offset

0,2

Comments

The Thue-Morse transform T acts on any binary word w with w(0) = 0 and in which both 0 and 1 appear infinitely often, by setting T(w)(0) = 0, T(w)(v_w(n)) = T(w)(n) and T(w)(u_w(n)) = 1 - T(w)(n), where v_w(n) and u_w(n) (offset 0) are the positions of the n-th 0 and the n-th 1 of w. Iterating T starting from a_0=A000035 produces A010060 (T(a_0)), A341389 (T^2(a_0)), the present sequence is T^3(a_0).

This sequence is 2-automatic. It is the fixed point starting with 0 of the uniform morphism of length 16 sending 0 -> 0110 0110 0110 0110 and 1 -> 1001 1001 1001 1001.

Links

Table of n, a(n) for n=0..89.

Benoit Cloitre, The Thue-Morse Transform, arXiv:2604.06243 [math.NT], 2026.

Formula

a(n) = Sum_{p >= 0, p AND 2 = 0} b_p(n) mod 2, where b_p(n) is the p-th binary digit of n.

a(2*n+1) = 1 - a(2*n) for n>=0.

a(16*n + r) = a(n) XOR a(r) for 0 <= r < 16 for n>=0.

Maple

a := proc(n) local s, p, bit;
    s := 0; p := 0;
    while 2^p <= n do
        if p mod 4 = 0 or p mod 4 = 1 then
            bit := iquo(n, 2^p) mod 2;
            s := (s + bit) mod 2 fi;
        p := p + 1;
    od; s end:
seq(a(n), n = 0..89);  # Peter Luschny, May 16 2026

Prog

(Python)
def a(n):
    s, p = 0, 0
    while (1 << p) <= n:
        if (p & 2) == 0:
            s ^= (n >> p) & 1
        p += 1
    return s
print([a(n) for n in range(90)])

Crossrefs

Cf. A000035, A010060, A341389, A395959, A395960, A395961, A395964.

Sequence in context: A219977 A128130 A133872 * A286903 A339052 A284490

Adjacent sequences: A395955 A395956 A395957 * A395959 A395960 A395961

Keyword

nonn,base,easy

Author

Benoit Cloitre, May 12 2026

Status

approved