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!
In this paper, the author gives the asymptotic formula
$$ P(n) = n^2 - 2n^{5/3} - 2n^{7/5} + \frac{13}{3}n^{4/3} - 2n^{9/7} + 2n^{6/5} - 2n^{13/11} + o(n^{13/11}) $$