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A145507 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=7.

%I A145507 #8 Jun 02 2025 00:39:45
%S A145507 7,61,3841,14760961,217885999165441,47474308632322991920487055361,
%T A145507 2253809980117057347661794063813616885861274573005652951041
%N A145507 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=7.
%C A145507 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 A145507 From Peter Bala, Nov 12 2012: (Start)
%F A145507 a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 4 + sqrt(15).
%F A145507 a(n) = 2*A005828(n) - 1.
%F A145507 Recurrence: a(n) = 9*{product {k = 1..n-1} a(k)} - 2 with a(1) = 7.
%F A145507 Product {n = 1..inf} (1 + 1/a(n)) = 3/10*sqrt(15).
%F A145507 Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(5/3).
%F A145507 (End)
%t A145507 aa = {}; k = 7; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
%t A145507 or
%t A145507 k =6; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)
%Y A145507 A145502-A145510. A005828.
%K A145507 nonn
%O A145507 1,1
%A A145507 _Artur Jasinski_, Oct 11 2008