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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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3
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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; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| 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 (paoloplava(AT)gmail.com), Aug 06 2009]
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LINKS
| Sergey Fomin and Andrei Zelevinsky, Cluster algebras II: Finite type classification
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FORMULA
| 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 (paoloplava(AT)gmail.com), Nov 27 2006
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MAPLE
| 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
| Cf. A076839, A076840, A076844.
Sequence in context: A171052 A021342 A069857 * A193088 A162916 A057273
Adjacent sequences: A076838 A076839 A076840 * A076842 A076843 A076844
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KEYWORD
| nonn
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Nov 21 2002
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