A145510 a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10

10, 118, 14158, 200477278, 40191139395243838, 1615327685887921300502934267457918, 2609283532796026943395592527806764363779539144932833602430435810558

Offset

1,1

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))

Links

Table of n, a(n) for n=1..7.

Formula

From Peter Bala, Nov 12 2012: (Start)

a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 1/2*(11 + sqrt(117)). a(n) = 1 (mod 9).

Recurrence: a(n) = 12*{Product_{k = 1..n-1} a(k)} - 2 with a(1) = 10.

Product {n = 1..inf} (1 + 1/a(n)) = 12/sqrt(117).

Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(13/9).

(End)

Mathematica

aa = {}; k = 10; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
(* or *)
k =9; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)

Crossrefs

Cf. A145502, A145503, A145504, A145505, A145506, A145507, A145508, A145509.

Sequence in context: A293987 A122887 A284331 * A160601 A338975 A327653

Adjacent sequences: A145507 A145508 A145509 * A145511 A145512 A145513

Keyword

nonn,easy

Author

Artur Jasinski, Oct 11 2008

Status

approved