

A037011


BaumSweet cubic sequence.


12



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
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OFFSET

1,1


COMMENTS

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


LINKS

Table of n, a(n) for n=1..92.
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. 118.]
Michael Gilleland, Some SelfSimilar Integer Sequences
H. Niederreiter and M. Vielhaber, Tree complexity and a doubly exponential gap between structured and random sequences, J. Complexity, 12 (1996), 187198.
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, n1, C(n1+k, n1k)*C(n1, k)) modulo 2 = A082759(n1) (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;


CROSSREFS

Cf. A086747, A106737, A277335.
Sequence in context: A014135 A014054 A014099 * A070563 A024692 A079978
Adjacent sequences: A037008 A037009 A037010 * A037012 A037013 A037014


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane.


STATUS

approved



