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k-2018-f-f-n-2n-f-k-2018-how-many-the-possible-of-k-integers-

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Question Number 32535 by naka3546 last updated on 27/Mar/18
k ≤ 2018 f (f (n) ) = 2n f (k) = 2018 how many the possible of k integers ?
k⩽2018f(f(n))=2nf(k)=2018howmanythepossibleofkintegers?

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