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Search: id:A119557
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| A119557 |
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a(1)=0,a(2)=0,a(3)=1 then a(n)=abs(a(n-1)-a(n-2))-a(n-3). |
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+0
2
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| 0, 0, 1, 1, 0, 0, -1, 1, 2, 2, -1, 1, 0, 2, 1, 1, -2, 2, 3, 3, -2, 2, 1, 3, 0, 2, -1, 3, 2, 2, -3, 3, 4, 4, -3, 3, 2, 4, -1, 3, 0, 4, 1, 3, -2, 4, 3, 3, -4, 4, 5, 5, -4, 4, 3, 5, -2, 4, 1, 5, 0, 4, -1, 5, 2, 4, -3, 5, 4, 4, -5, 5, 6, 6, -5, 5, 4, 6, -3, 5, 2, 6, -1, 5, 0, 6, 1, 5, -2, 6, 3, 5, -4, 6, 5, 5, -6, 6, 7, 7
(list; graph; listen)
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OFFSET
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0,9
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COMMENT
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See A104156 for an order 2 example.
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REFERENCES
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B. Cloitre, On strange predictible recursions, preprint 2006
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FORMULA
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abs(a(2n-1))=A004738(n)-1 where sign(a(2*n-1)) alternates between 2 consecutive zeros. b(n)=a(2n)-floor(sqrt(n))+1 is an infinite binary word consisting in a sequence of bloc (0,1) and single 0's where 0's occur when n is of form k^2-1, k>=2 i.e. b(n) begins for n>=1 : (0,1),0,(0,1),(0,1),0,(0,1),(0,1),(0,1),0,(0,1),... and single 0's occur at n=3,8,15,...
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PROGRAM
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(PARI) ? an=vector(10000); ? an[1]=0; an[2]=0; an[3]=1; ? a(n)=if(n<0, 0, an[n]); ? for(n=4, 10000, an[n]=abs(a(n-1)-a(n-2))-a(n-3))
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CROSSREFS
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Cf. A104156.
Sequence in context: A081389 A133685 A112183 this_sequence A125919 A061198 A039801
Adjacent sequences: A119554 A119555 A119556 this_sequence A119558 A119559 A119560
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KEYWORD
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sign
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), May 30 2006
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