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 A073089 (1/2)*(4n - 3 - Sum (k=1..n, A007400(k))). 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS From Joerg Arndt, Oct 28 2013: (Start) Sequence is (essentially) obtained by complementing every other term of A014577. Turns (by 90 degrees) of a curve similar to the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw). See the linked pdf files for two renderings of the curve. (End) LINKS Joerg Arndt, pdf rendering of the curve described in comment, alternate rendering of the curve. FORMULA 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 EXAMPLE From Paul D. Hanna, Oct 19 2012: (Start) 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.  (End) PROG (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])) \\ 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 (PARI) a(n) = { if ( n<=1, return(0)); n-=1; my(v=2^valuation(n, 2) ); return( (0==bitand(n, v<<1)) != (v%2) ); } \\ Joerg Arndt, Oct 28 2013 CROSSREFS Cf. A007400. Sequence in context: A014306 A138150 A271591 * A011657 A072126 A111113 Adjacent sequences:  A073086 A073087 A073088 * A073090 A073091 A073092 KEYWORD easy,nonn AUTHOR Benoit Cloitre, Aug 18 2002 STATUS approved

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