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The number of prime powers (exponents $\geq$ 2) up to x is given by:

$x^\left(\frac12\right)+x^\left(\frac13\right)+x^\left(\frac14\right)+ $...$ =O(\sqrt x$ $lnx) $

(http://mathworld.wolfram.com/PrimePower.html)

I am not sure of the $O(...)$ here, and is couting prime powers with exponents $\geq 2$ this straightforward? Is the equation above an approximation / what should be done to get the exact prime power count?

Navneeth G
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    Error in the formula reported to mathworld.com. – gnasher729 Oct 27 '14 at 16:16
  • 6 years later, still not fixed. Clearly that is counting powers, rather than prime powers. We want to sum $\pi(x^{1/2})+\pi(x^{1/3})+...$ where $\pi(x)$ is the prime counting function. Also, yes it is that simple, start with 1 rather than 2, so we count the primes with exponent 1, and that does get you the exact prime power count. – DanaJ Mar 05 '21 at 06:25

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That looks very wrong to me.

The number of primes ≤ n is about $n / \ln n$. The number of prime squares ≤ n is exactly the same as the number of primes ≤ $n^{1/2}$ which is $2 n^{1/2} / \ln n$, and so on.

The number of cubes, fourth powers etc. is quite negligible, so the number of prime powers ≤ x with exponent ≥ 2 is about $2 n^{1/2} / \ln n$.

gnasher729
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