A395961 Fourth iterate of the Thue-Morse transform applied to A000035.

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

Offset

0,1

Comments

The Thue-Morse transform T acts on any binary word w with w(0) = 0 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)), A395958 (T^3(a_0)), the present sequence is T^4(a_0).

This sequence is 2-automatic. It is the fixed point starting with 0 of the uniform morphism of length 16 sending 0 -> 0101 0101 0101 0101 and 1 -> 1010 1010 1010 1010.

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 3 = 0} b_p(n) mod 2, where b_p(n) is the p-th binary digit of n.

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

a(16n + r) = a(n) XOR (r mod 2) for 0 <= r < 16 and n >= 0.

Prog

(Python)
def a(n):
    s, p = 0, 0
    while (1 << p) <= n:
        if (p & 3) == 0:
            s ^= (n >> p) & 1
        p += 1
    return s
print([a(n) for n in range(90)])

Crossrefs

Cf. A000035, A010060, A341389, A395958, A395959, A395960, A395962, A395963, A395964.

Sequence in context: A091337 A131734 A134452 * A327515 A327532 A073445

Adjacent sequences: A395958 A395959 A395960 * A395962 A395963 A395964

Keyword

nonn,base,easy

Author

Benoit Cloitre, May 12 2026

Status

approved