A073089 a(n) = (1/2)*(4n - 3 - Sum_{k=1..n} A007400(k)).

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

Offset: 1

Author

Benoit Cloitre, Aug 18 2002

Keywords

• easy
• nonn

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)

Examples

			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)
		

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.
• Index entries for characteristic functions

Crossrefs

Cf. A007400, A073088 (the sum part here), A123725.

Programs

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

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