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%I #47 Mar 13 2021 00:11:47
%S 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,
%T 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,
%U 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
%N a(n) = (1/2)*(4n - 3 - Sum_{k=1..n} A007400(k)).
%C From _Joerg Arndt_, Oct 28 2013: (Start)
%C Sequence is (essentially) obtained by complementing every other term of A014577.
%C 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).
%C See the linked pdf files for two renderings of the curve. (End)
%H Antti Karttunen, <a href="/A073089/b073089.txt">Table of n, a(n) for n = 1..65537</a>
%H Joerg Arndt, pdf rendering of the <a href="/A073089/a073089.pdf">curve described in comment</a>, <a href="/A073089/a073089_1.pdf">alternate rendering of the curve</a>.
%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>
%F 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).
%F G.f.: The following series has a simple continued fraction expansion:
%F 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
%F 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
%e From _Paul D. Hanna_, Oct 19 2012: (Start)
%e Let F(x) = x + 1/x + 1/x^3 + 1/x^7 + 1/x^15 + 1/x^31 +...+ 1/x^(2^n-1) +...
%e 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 +...)))))))))))))))))))))),
%e a continued fraction in which the partial quotients equal (-1)^a(n)*x. (End)
%o (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_
%o (PARI) /* Using the Continued Fraction, Print 2^N terms of this sequence: */
%o {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
%o (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
%Y Cf. A007400, A073088 (the sum part here), A123725.
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, Aug 18 2002
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