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A145510 a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10.

%I A145510 #12 Jun 02 2025 00:40:06
%S A145510 10,118,14158,200477278,40191139395243838,
%T A145510 1615327685887921300502934267457918,
%U A145510 2609283532796026943395592527806764363779539144932833602430435810558
%N A145510 a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10.
%C A145510 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 A145510 From _Peter Bala_, Nov 12 2012: (Start)
%F A145510 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 A145510 Recurrence: a(n) = 12*{Product_{k = 1..n-1} a(k)} - 2 with a(1) = 10.
%F A145510 Product {n = 1..inf} (1 + 1/a(n)) = 12/sqrt(117).
%F A145510 Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(13/9).
%F A145510 (End)
%t A145510 aa = {}; k = 10; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
%t A145510 (* or *)
%t A145510 k =9; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)
%Y A145510 Cf. A145502, A145503, A145504, A145505, A145506, A145507, A145508, A145509.
%K A145510 nonn,easy
%O A145510 1,1
%A A145510 _Artur Jasinski_, Oct 11 2008