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%I #31 Jan 07 2023 04:04:23
%S 2,6,46,2206,4870846,23725150497406,562882766124611619513723646,
%T 316837008400094222150776738483768236006420971486980606
%N a(n+1) = a(n)^2+2*a(n)-2 and a(1)=2.
%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))).
%C From _Peter Bala_, Nov 12 2012: (Start)
%C The present sequence corresponds to the case x = 3 of the following general remarks. Sequences A145503 through A145510 correspond to the cases x = 4 through x = 11 respectively.
%C Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1. Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^2 + 2*a(n) - 2 with the initial condition a(1) = x - 1.
%C A second recurrence is a(n) = (a(1) + 2)*{Product_{k = 1..n-1} a(k)} - 2.
%C The following algebraic identity is valid for x > 2:
%C (x + 1)/sqrt(x^2 - 4) = (1 + 1/(x - 1))*(y + 1)/sqrt(y^2 - 4), where y - 1 = (x - 1)^2 + 2*(x - 1) - 2. Iterating the identity yields the product expansion (x + 1)/sqrt(x^2 - 4) = Product {n = 1..oo} (1 + 1/a(n)).
%C A second expansion is Product {n = 1..oo} (1 + 2/(a(n) + 1) = sqrt((x + 2)/(x - 2)). For an alternative approach to these identities see the Bala link.
%C (End)
%H Peter Bala, <a href="/A145502/a145502_1.pdf">Notes on A145502-A145510</a>.
%H Daniel Duverney, <a href="https://www.ms.u-tokyo.ac.jp/journal/abstract/jms080206.html">Irrationality of Fast Converging Series of Rational Numbers</a>, Journal of Mathematical Sciences-University of Tokyo, Vol. 8, No. 2 (2001), pp. 275-316.
%H Daniel Duverney and Takeshi Kurosawa, <a href="https://doi.org/10.1007/s40993-022-00366-1">Transcendence of infinite products involving Fibonacci and Lucas numbers</a>, Research in Number Theory, Vol. 8 (2002), Article 68.
%F From _Peter Bala_, Nov 12 2012: (Start)
%F a(n) = phi^(2^n) + (1/phi)^(2^n) - 1, where phi := (1 + sqrt(5))/2 is the golden ratio.
%F a(n) = A001566(n-1) - 1.
%F Recurrence: a(n) = 4*(Product_{k = 1..n-1} a(k)) - 2 with a(1) = 2.
%F Product_{n >= 1} (1 + 1/a(n)) = 4/sqrt(5).
%F Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt(5). (End)
%F From _Amiram Eldar_, Sep 10 2022: (Start)
%F a(n) = A000324(n) - 3.
%F Sum_{n>=1} (-2)^n/a(n) = -1/2 (Duverney, 2001). (End)
%F Product_{n>=1} (1 + 3/a(n)) = 4 (Duverney and Kurosawa, 2022). - _Amiram Eldar_, Jan 07 2023
%t aa = {}; k = 2; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
%t (* or *)
%t k = 1; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}]
%t NestList[#^2+2#-2&,2,10] (* _Harvey P. Dale_, Dec 14 2021 *)
%Y Cf. A145503, A145504, A145505, A145506, A145507, A145508, A145509, A145510.
%Y Cf. A000324, A001566.
%K nonn,easy
%O 1,1
%A _Artur Jasinski_, Oct 11 2008
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