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 A073089 a(n) = (1/2)*(4n - 3 - Sum_{k=1..n} A007400(k)). 5
 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 Antti Karttunen, Table of n, a(n) for n = 1..65537 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 a(n) = A014577(n-2) + A056594(n).  Conjecture: a(n) = (1 + (-1)^A057661(n - 1))/2 for all n > 1. - Velin Yanev, Feb 01 2021 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, A073088 (the sum part here), A123725. Sequence in context: A271591 A287790 A285159 * A323158 A011657 A072126 Adjacent sequences:  A073086 A073087 A073088 * A073090 A073091 A073092 KEYWORD easy,nonn AUTHOR Benoit Cloitre, Aug 18 2002 STATUS approved

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Last modified December 7 12:17 EST 2021. Contains 349581 sequences. (Running on oeis4.)