A145508 - OEIS

%I #10 Sep 20 2013 15:31:15

%S 8,78,6238,38925118,1515164889164158,2295724641355838227053650177278,

%T 5270351628928392053240255925779522360603268430188068127843838

%N a(n+1)=a(n)^2+2*a(n)-2 and a(1)=8

%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*(9 + sqrt(77)). a(n) = 1 (mod 7).

%F Recurrence: a(n) = 10*{product {k = 1..n-1} a(k)} - 2 with a(1) = 8.

%F Product {n = 1..inf} (1 + 1/a(n)) = 10/sqrt(77).

%F Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(11/7).

%F (End)

%t aa = {}; k = 8; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa

%t or

%t k =7; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)

%t NestList[#^2+2#-2&,8,8] (* _Harvey P. Dale_, Sep 20 2013 *)

%Y A145502-A145510.

%K nonn,easy

%O 1,1

%A _Artur Jasinski_, Oct 11 2008