A145507 - OEIS

%I #6 Nov 12 2012 07:19:04

%S 7,61,3841,14760961,217885999165441,47474308632322991920487055361,

%T 2253809980117057347661794063813616885861274573005652951041

%N a(n+1)=a(n)^2+2*a(n)-2 and a(1)=7

%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 := 4 + sqrt(15).

%F a(n) = 2*A005828(n) - 1.

%F Recurrence: a(n) = 9*{product {k = 1..n-1} a(k)} - 2 with a(1) = 7.

%F Product {n = 1..inf} (1 + 1/a(n)) = 3/10*sqrt(15).

%F Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(5/3).

%F (End)

%t aa = {}; k = 7; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa

%t or

%t k =6; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)

%Y A145502-A145510. A005828.

%K nonn

%O 1,1

%A _Artur Jasinski_, Oct 11 2008