What is the $n^\text{th}$ perfect power, $P(n)$?

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!

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

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