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!

  • 6
    $\begingroup$ 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


In this paper, the author gives the asymptotic formula

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.