

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
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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

Table of n, a(n) for n=1..105.
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^n1) = [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^n1) +...
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^n1)), 2^N); for(n=1, 2^N, print1((1CF[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: A165560 A014306 A138150 * A011657 A072126 A111113
Adjacent sequences: A073086 A073087 A073088 * A073090 A073091 A073092


KEYWORD

easy,nonn


AUTHOR

Benoit Cloitre, Aug 18 2002


STATUS

approved



