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Search: id:A037011
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| A037011 |
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Baum-Sweet cubic sequence. |
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+0
10
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| 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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Memo: more sequences like this should be added to the database.
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REFERENCES
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H. Niederreiter and M. Vielhaber, Tree complexity and a doubly ..., J. Complexity, 12 (1996), 187-198.
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LINKS
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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
D. P. Robbins, Cubic Laurent series in characteristic 2 with bounded partial quotients
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FORMULA
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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 (benoit7848c(AT)orange.fr), Jun 20 2003
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MAPLE
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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;
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CROSSREFS
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Cf. A086747.
Adjacent sequences: A037008 A037009 A037010 this_sequence A037012 A037013 A037014
Sequence in context: A014135 A014054 A014099 this_sequence A024692 A079978 A068429
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KEYWORD
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nonn,easy
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AUTHOR
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njas
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