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

7, 61, 3841, 14760961, 217885999165441, 47474308632322991920487055361, 2253809980117057347661794063813616885861274573005652951041

Offset: 1

Author

Artur Jasinski, Oct 11 2008

Keywords

• nonn

Comments

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))

Crossrefs

A145502-A145510. A005828.

Programs

Mathematica

aa = {}; k = 7; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
or
k =6; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)

Formula

From Peter Bala, Nov 12 2012: (Start)

a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 4 + sqrt(15).

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

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

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

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

(End)