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Let $P(n)$ denotes the $n^\text{th}$ perfect power of natural numbers (in ascending order without repetition).

So, $P(1)=1, P(2)=4, P(3)=8, P(4)=9, P(5)=16, P(6)=25, P(7)=27, P(8)=32, \dots$.

Is there any formula to find $P(n)$ for a natural number $n$? Say what is $P(75)$?

I think there is no explicit formula for $P(n)$ since the sequence $P(1), P(2), P(3), \dots$ has no common difference of any order, has no common ratio, and has no pattern in the slopes.

What I mean is; if we sketch the line graph $y=P(x)$, then $P'(x)$ at $x=n$ is always positive, but not always increasing nor always decreasing.

Any help to find the $n^\text{th}$ perfect power would be appreciated. THANKS!

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    $\begingroup$ oeis.org/A001597 $\endgroup$
    – dan_fulea
    Sep 3, 2019 at 1:09
  • $\begingroup$ it's always at least $2^x$ when $n\geq T(x)$ where the function is the x-th triangular number. $\endgroup$
    – user645636
    Sep 3, 2019 at 11:15

1 Answer 1

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In this paper, the author gives the asymptotic formula

$$ P(n) = n^2 - 2n^{5/3} - 2n^{7/5} + \frac{13}{3}n^{4/3} - 2n^{9/7} + 2n^{6/5} - 2n^{13/11} + o(n^{13/11}) $$

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