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Let x(n) denote the number of 4's among the n first elements of the continued fraction for sum k>=0 1/2^(2^k) (A007400 ), y(n) the number of 6's and z(n) the number of 2's. Then a(n)=x(n)-y(n)-z(n)-1.

+0
4

-1, -1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, -1, 0, -1 (list; graph; listen)

	OFFSET	`0,1`
	COMMENT	`The positive sequence has a(n)=mod(A000120(A047849(n)),2)=mod(A000120(A078008(2n)),2) - Paul Barry (pbarry(AT)wit.ie), Jan 13 2005` `Cosh(1) in 'reflected factorial' base is 1.10101010101010101010101010101010101010101010... - see A091337 for Sinh(1) (from Robert G. Wilson v (rgwv(AT)rgwv.com), May 04 2005)`
	FORMULA	`It seems that a(2k+1)=0 for k>=1.` `The positive sequence (assuming the pattern continues) has g.f. (1+x-x^2)/((1-x)(1-x^2)), with a(n)=(1-(1)^n)/2+0^n=mod((1+A001045(n+1))/2, 2) =mod(A005578, 2). The partial sums are A008619(n+1). - Paul Barry (pbarry(AT)wit.ie), Apr 28 2004`
	CROSSREFS	`Cf. A007400.` `Adjacent sequences: A073094 A073095 A073096 this_sequence A073098 A073099 A073100` `Sequence in context: A068427 A080545 A141735 this_sequence A117569 A135528 A071040`
	KEYWORD	`sign`
	AUTHOR	`Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 18 2002`

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Last modified October 13 20:18 EDT 2008. Contains 145016 sequences.

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