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A073097
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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.
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3
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-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; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| 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)
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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
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CROSSREFS
| Cf. A007400.
Sequence in context: A080545 A141735 A178225 * A117569 A135528 A163805
Adjacent sequences: A073094 A073095 A073096 * A073098 A073099 A073100
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KEYWORD
| sign
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 18 2002
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