A037011 Baum-Sweet cubic sequence.

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

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Memo: more sequences like this should be added to the database.

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• J.-P. Allouche, Finite automata and arithmetic Seminaire Lotharingien de Combinatoire, B30c (1993), 23 pp. [Formerly: Publ. I.R.M.A. Strasbourg, 1993, 1993/034, p. 1-18.]
• Michael Gilleland, Some Self-Similar Integer Sequences
• H. Niederreiter and M. Vielhaber, Tree complexity and a doubly exponential gap between structured and random sequences, J. Complexity, 12 (1996), 187-198.
• D. P. Robbins, Cubic Laurent series in characteristic 2 with bounded partial quotients, arXiv:math/9903092 [math.NT], 1999.
• Index entries for characteristic functions

Crossrefs

Cf. A086747, A106737, A277335.

Programs

Maple

A := x; for n from 1 to 100 do series(x+x*A^3+O(x^(n+2)),x,n+2); A := series(% mod 2,x,n+2); od: A;

Mathematica

m = 100; A[_] = 0;
Do[A[x_] = x + x A[x]^3 + O[x]^m // Normal // PolynomialMod[#, 2]&, {m}];
CoefficientList[A[x], x] // Rest (* Jean-François Alcover, Oct 15 2019 *)

Formula

G.f. satisfies A^3+x^(-1)*A+1 = 0 (mod 2).

It appears that a(n)=sum(k=0, n-1, C(n-1+k, n-1-k)*C(n-1, k)) modulo 2 = A082759(n-1) (mod 2). It appears also that a(k)=1 iff k/3 is in A003714. - Benoit Cloitre, Jun 20 2003

From Antti Karttunen, Nov 03 2017: (Start)

If Cloitre's above observation holds, then we also have (assuming starting offset 0, with a(0) = 1):

a(n) = A000035(A106737(n)).

a(n) = A010052(A005940(1+n)).

(End)