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%I #10 Mar 30 2020 08:43:23
%S 10,118,14158,200477278,40191139395243838,
%T 1615327685887921300502934267457918,
%U 2609283532796026943395592527806764363779539144932833602430435810558
%N a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10
%C General formula for a(n+1)=a(n)^2+2*a(n)-2 and a(1)=k+1 is a(n)=Floor[((k + Sqrt[k^2 + 4])/2)^(2^((n+1) - 1))
%F From _Peter Bala_, Nov 12 2012: (Start)
%F a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 1/2*(11 + sqrt(117)). a(n) = 1 (mod 9).
%F Recurrence: a(n) = 12*{Product_{k = 1..n-1} a(k)} - 2 with a(1) = 10.
%F Product {n = 1..inf} (1 + 1/a(n)) = 12/sqrt(117).
%F Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(13/9).
%F (End)
%t aa = {}; k = 10; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
%t (* or *)
%t k =9; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)
%Y Cf. A145502, A145503, A145504, A145505, A145506, A145507, A145508, A145509.
%K nonn,easy
%O 1,1
%A _Artur Jasinski_, Oct 11 2008
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