A395088 Fixed point of the morphism 0->011, 1->0.

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

Offset

0

Comments

Start with 0, apply 0 -> 011, 1 -> 0, take limit.

Define strings S(0) = 1, S(1) = 0, S(n) = S(n-1)S(n-2)S(n-2); iterate; sequence is S(infinity).

Related to the Thue-Morse sequence A010060 (see formula section).

Conjecture: Let c(n) be the list with the length of 0-runs, then a(n) = (3-c(n+2))/2, i.e., c(1) = 1, c(n) = 3 - 2*a(n-2) for n > 1. If this conjecture is true, this is a self-generating sequence, starting with 011 whereafter the n-th 0-run has length 3 - 2*a(n-2) and the 1-runs all have length 2.

Links

Paolo Xausa, Table of n, a(n) for n = 0..10922

Formula

a(n) = 1 - A395089(n).

a(n) = 2 - A026465(n+2).

P(a(n) = 0) = P(a(n) = 1) = 1/2.

Example

011000110110110001100011000110110110001101101100011...

Mathematica

f[n_]:=Module[{a={0}, b={1}}, While[Length[a]<n, {a, b}={Join[a, b, b], a}]; Take[a, n]];
a[n_]:=Module[{a={0}, b={1}}, While[Length[a]<=n, {a, b}={Join[a, b, b], a}]; a[[n+1]]];
a/@Range[0, 86] (* Robert P. P. McKone, Apr 14 2026 *)
(* Alternative: *)
First[SubstitutionSystem[{0 -> {0, 1, 1}, 1 -> {0}}, {0}, {7}]] (* Paolo Xausa, May 04 2026 *)

Prog

(Python)
def first(n): # returns first n terms
    f0, f1 = "01"
    while len(f0) < n:
        f0, f1 = f0+2*f1, f0
    return list(map(int, f0[0:n]))
print(first(87))

Crossrefs

Cf. A010060, A026465.

Cf. A003849, A005614, A171588, A080764, A395089.

Sequence in context: A254114 A105384 A288694 * A292077 A327256 A327177

Adjacent sequences: A395085 A395086 A395087 * A395089 A395090 A395091

Keyword

nonn,easy

Author

A.H.M. Smeets, Apr 11 2026

Status

approved