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A037011
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Baum-Sweet cubic sequence.
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13
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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, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0
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OFFSET
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1,1
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COMMENTS
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Memo: more sequences like this should be added to the database.
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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.]
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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, Jun 20 2003
If Cloitre's above observation holds, then we also have (assuming starting offset 0, with a(0) = 1):
(End)
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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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MATHEMATICA
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m = 100; A[_] = 0;
Do[A[x_] = x + x A[x]^3 + O[x]^m // Normal // PolynomialMod[#, 2]&, {m}];
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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