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 $$x_1 = \sqrt[3]{3},\; x_{n+1}=(x_n)^{\sqrt[3]{3}}$$. Find smallest $$n$$ such that $$x_n$$ is integer
02-20-2010, 02:52 PM
Post: #1
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
$$x_1 = \sqrt[3]{3},\; x_{n+1}=(x_n)^{\sqrt[3]{3}}$$. Find smallest $$n$$ such that $$x_n$$ is integer
By indection, one can easily verify that
$x_n = 3^{3^{-1+(n-1)/3}}$ thus $$n = 4$$ is the answer.
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