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

1,1

Comments

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

Links

Table of n, a(n) for n=1..105.

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

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)

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 *)

Crossrefs

Cf. A086747, A106737, A277335.

Sequence in context: A014135 A014054 A014099 * A070563 A374053 A024692

Adjacent sequences: A037008 A037009 A037010 * A037012 A037013 A037014

Keyword

nonn,easy

Author

N. J. A. Sloane.

Status

approved