Given that y equals f (x) is an increasing function, for any n subordinate to a natural number, f [f (n)] equals 3N, find f (1) + F (6) + F (18)

Given that y equals f (x) is an increasing function, for any n subordinate to a natural number, f [f (n)] equals 3N, find f (1) + F (6) + F (18)

Y = f (x), n ∈ n + is a (strictly) increasing function, for any n ∈ n + there is f [f (n)] = 3N ∧ for any n ∈ n + there is f (n) ∈ n ∫ f (n)) = 3N, ∧ f (1)) = 3, if f (1) = 1, then f (1)) = f (1) = 1 ∧ f (1) ≠ 1 ∧ f (1) ≥ 2 ∧ f (x) is an increasing function ∧ f (2) ≤ f (1)) = 3 ∧ f (3) ≥ f (f (2))

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