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Search: id:A076841
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| A076841 |
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a(1) = a(2) = 1; a(n) = (a(n-1)+1)/a(n-2) (for n>2, n odd), (a(n-1)^3+1)/a(n-2) (for n>2, n even). |
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4
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| 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Any sequence a(1),a(2),a(3),... defined by the recurrence a(n) = (a(n-1)+1)/a(n-2) (for n>2, n odd), (a(n-1)^3+1)/a(n-2) (for n>2, n even) has period 8. The theory of cluster algebras currently being developed by Fomin and Zelevinsky gives a context for these facts, but it doesn't really explain them in an elementary way. - James Propp, Nov 20, 2002
Terms of the simple continued fraction of 3487/[4*sqrt(1621590)-3016]. [From Paolo P. Lava (ppl(AT)spl.at), Aug 06 2009]
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LINKS
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Sergey Fomin and Andrei Zelevinsky, Cluster algebras II: Finite type classification
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FORMULA
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a(n)=1/224*{65*(n mod 8)+65*[(n+1) mod 8]+345*[(n+2) mod 8]-215*[(n+3) mod 8]+149*[(n+4) mod 8]-159*[(n+5) mod 8]+9*[(n+6) mod 8]+37*[(n+7) mod 8]} with n>=0 - Paolo P. Lava (ppl(AT)spl.at), Nov 27 2006
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MAPLE
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a := 1; b := 1; f := proc(n) option remember; global a, b; if n=1 then RETURN(a); fi; if n=2 then RETURN(b); fi; if n mod 2 = 1 then RETURN((f(n-1)+1)/f(n-2)); fi; RETURN((f(n-1)^3+1)/f(n-2)); end;
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CROSSREFS
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Cf. A076839, A076840, A076844.
Sequence in context: A051491 A021342 A069857 this_sequence A162916 A057273 A108986
Adjacent sequences: A076838 A076839 A076840 this_sequence A076842 A076843 A076844
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Nov 21 2002
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