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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)^2 + 1)/a(n-2) (for n>2, n even).
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%I #24 Feb 24 2024 01:04:39

%S 1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,

%T 3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,

%U 2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2,5,3,2,1,1,2

%N a(1) = a(2) = 1; a(n) = (a(n-1) + 1)/a(n-2) (for n>2, n odd), (a(n-1)^2 + 1)/a(n-2) (for n>2, n even).

%C 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)^2+1)/a(n-2) (for n>2, n even) has period 6. 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

%H Sergey Fomin and Andrei Zelevinsky, <a href="https://arxiv.org/abs/math/0208229">Cluster algebras II: Finite type classification</a>, arXiv:math/0208229 [math.RA], 2002-2003.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,1).

%p 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)^2+1)/f(n-2)); end;

%t LinearRecurrence[{0, 0, 0, 0, 0, 1}, {1, 1, 2, 5, 3, 2}, 105] (* _Jean-François Alcover_, Nov 22 2017 *)

%Y Cf. A076839, A076841, A076844.

%K nonn,easy

%O 1,3

%A _N. J. A. Sloane_, Nov 21 2002