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A073089
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(1/2)*(4n - 3 - Sum (k=1,n, A007400(k))).
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3
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0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1
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OFFSET
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1,1
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LINKS
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Table of n, a(n) for n=1..105.
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FORMULA
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Recurrence: a(1) = a(4n+2) = a(8n+7) = a(16n+13) = 0, a(4n) = a(8n+3) = a(16n+5) = 1, a(8n+1) = a(4n+1).
G.f.: The following series has a simple continued fraction expansion:
x + Sum_{n>=1} 1/x^(2^n-1) = [x; x, -x, -x, -x, x, ..., (-1)^a(n)*x, ...]. - Paul D. Hanna, Oct 19 2012
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EXAMPLE
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Let F(x) = x + 1/x + 1/x^3 + 1/x^7 + 1/x^15 + 1/x^31 +...+ 1/x^(2^n-1) +...
then F(x) = x + 1/(x + 1/(-x + 1/(-x + 1/(-x + 1/(x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(-x + 1/(-x + 1/(-x + 1/(x +...+ 1/((-1)^a(n)*x +...)))))))))))))))))))))),
a continued fraction in which the partial quotients equal (-1)^a(n)*x. - Paul D. Hanna, Oct 19 2012
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PROG
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(PARI) a(n)=if(n<2, 0, if(n%8==1, a((n+1)/2), [1, -1, 0, 1, 1, 1, 0, 0, 1, -1, 0, 1, 1, 0, 0, 0][(n%16)+1])) /* from Ralf Stephan */
(PARI) /* Using the Continued Fraction, Print 2^N terms of this sequence: */
{N=10; CF=contfrac(x+sum(n=1, N, 1/x^(2^n-1)), 2^N); for(n=1, 2^N, print1((1-CF[n]/x)/2, ", "))} \\ Paul D. Hanna, Oct 19 2012
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CROSSREFS
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Cf. A007400.
Sequence in context: A165560 A014306 A138150 * A011657 A072126 A111113
Adjacent sequences: A073086 A073087 A073088 * A073090 A073091 A073092
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KEYWORD
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easy,nonn,changed
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AUTHOR
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Benoit Cloitre, Aug 18 2002
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STATUS
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approved
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