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)).
Essentially the same as A110407. [R. J. Mathar, Mar 18 2009]
LINKS
Indranil Ghosh, Table of n, a(n) for n = 1..11
FORMULA
From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 2 + sqrt(3).
Recurrence: a(n) = 5*(Product {k = 1..n-1} a(k)) - 2 with a(1) = 3.
Product_{n >= 1} (1 + 1/a(n)) = 5/6*sqrt(3).
Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt(3).
(End)
MATHEMATICA
aa = {}; k = 3; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
(* or *)
k = 2; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}]
NestList[#^2+2#-2&, 3, 10] (* Harvey P. Dale, Feb 01 2018 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Artur Jasinski, Oct 11 2008
STATUS
approved