|
|
A145503
|
|
a(n+1) = a(n)^2+2*a(n)-2 and a(1)=3.
|
|
5
|
|
|
|
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
|
|
|
FORMULA
|
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}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|