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A145509
a(n+1)=a(n)^2+2*a(n)-2 and a(1)=9
2
9, 97, 9601, 92198401, 8500545331353601, 72259270930397519221389558374401, 5221402235392591963136699520829303150191924374488750728808857601
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))
FORMULA
From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 5 + 2*sqrt(6). a(n) = 1 (mod 8).
Recurrence: a(n) = 11*{product {k = 1..n-1} a(k)} - 2 with a(1) = 9.
Product {n = 1..inf} (1 + 1/a(n)) = 11/sqrt(96).
Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(3/2).
(End)
MATHEMATICA
aa = {}; k = 9; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
or
k =8; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)
NestList[#^2+2#-2&, 9, 10] (* Harvey P. Dale, Jul 02 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Artur Jasinski, Oct 11 2008
STATUS
approved