A395964 Second iterate of the Thue-Morse transform applied to the Fibonacci word A003849.

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

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.

Applied to the Fibonacci word f = A003849 (fixed point of 0 -> 01, 1 -> 0), one has T(f) = A095076 (Fibonacci-Thue-Morse). The present sequence is T(T(f)) = T(A095076).

In contrast to the dyadic iterates A010060, A341389, A395958, A395961, this sequence is not 2-automatic.

The positions of 0 are given by A395965 and the positions of 1 by A395966.

Links

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

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

Prog

(Python)
def fib_word(L):
    s = [0]
    while len(s) < L:
        s = sum(([0, 1] if c == 0 else [0] for c in s), [])
    return s[:L]
def T(w):
    N = len(w)
    v = [i for i, b in enumerate(w) if b == 0]
    u = [i for i, b in enumerate(w) if b == 1]
    wp = [None] * N
    wp[0] = 0
    for k in range(N):
        if wp[k] is None: break
        if k < len(v) and v[k] < N: wp[v[k]] = wp[k]
        if k < len(u) and u[k] < N: wp[u[k]] = 1 - wp[k]
    return wp
print(T(T(fib_word(200)))[:90])

Crossrefs

Cf. A003849, A010060, A095076, A341389, A395958, A395961, A395965, A395966.

Sequence in context: A090171 A316832 A086747 * A188973 A188970 A244992

Adjacent sequences: A395961 A395962 A395963 * A395965 A395966 A395967

Keyword

nonn

Author

Benoit Cloitre, May 12 2026

Status

approved