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 A076841 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). 3
 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; text; internal format)
 OFFSET 1,3 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, Aug 06 2009] LINKS Sergey Fomin and Andrei Zelevinsky, Cluster algebras II: Finite type classification Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 1). 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, Nov 27 2006 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; MATHEMATICA LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 1}, {1, 1, 2, 9, 5, 14, 3, 2}, 99] (* Ray Chandler, Aug 25 2015 *) CROSSREFS Cf. A076839, A076840, A076844. Sequence in context: A171052 A021342 A069857 * A213819 A193088 A162916 Adjacent sequences:  A076838 A076839 A076840 * A076842 A076843 A076844 KEYWORD nonn AUTHOR N. J. A. Sloane, Nov 21 2002 STATUS approved

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Last modified October 16 03:37 EDT 2019. Contains 328040 sequences. (Running on oeis4.)