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a(n+1)=a(n)^2+2*a(n)-2 and a(1)=4
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%I #9 Nov 16 2013 18:25:52

%S 4,22,526,277726,77132286526,5949389624883225721726,

%T 35395236908668169265765137996816180039862526,

%U 1252822795820745419377249396736955608088527968701950139470082687906021780162741058825726

%N a(n+1)=a(n)^2+2*a(n)-2 and a(1)=4

%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 The next term (a(9)) has 175 digits. - _Harvey P. Dale_, Nov 16 2013

%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*(5 + sqrt(21)).

%F a(n) = A003487(n-1) - 1.

%F Recurrence: a(n) = 6*{product {k = 1..n-1} a(k)} - 2 with a(1) = 4.

%F Product {n = 1..inf} (1 + 1/a(n)) = 2/7*sqrt(21).

%F Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(7/3).

%F (End)

%t NestList[#^2+2#-2&,4,7] (* _Harvey P. Dale_, Nov 16 2013 *)

%Y A145502-A145510. A003487.

%K nonn

%O 1,1

%A _Artur Jasinski_, Oct 11 2008

%E One additional term (a(8)) from _Harvey P. Dale_, Nov 16 2013