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Is the set of numbers that are NOT prime powers infinite?

I know there is an infinite amount of primes...and therefore an infinite amount of prime powers.

I'm just curious if the set if those that are nit prime powers is also infinite.

Thanks!

Ethan Bolker
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Jon Saboe
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    The set ${6^n}_{n=1}^{\infty}$ is already infinite. – lulu Jun 15 '18 at 15:23
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    Pick any prime $p > 2$ and double it. $2p$ is not a prime power. – gandalf61 Jun 15 '18 at 15:24
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    Choose two distinct primes and take the product. This is clearly not a prime power, and since there are infinite many primes, you also have infinite many such pairs. – Peter Jun 16 '18 at 08:39
  • The easiest infinite sequence (with respect to the decimal expansion) of non-prime-powers is the sequence of the powers of $10$. A $1$ followed by an arbitary number of zeros is never a prime power. – Peter Jun 16 '18 at 08:47
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    Most numbers are not prime powers. The number of prime powers [including primes] not exceeding $x$ is asymptotically equal to $\operatorname{Li}(x)$ (or, if you want a simpler approximating function that however gives a slightly worse approximation, to $\frac{x}{\log x}$). Almost all of those are primes. The number of prime powers not exceeding $x$ excluding primes is $O(\sqrt{x})$. – Daniel Fischer Jun 16 '18 at 10:50

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Of course the set of numbers that are not prime powers is infinite.

If I'm understanding your question correctly, what you're looking for are the squarefree composite numbers. Ten thousand of them are listed in Sloane's A120944, and the only reason it doesn't list more is because the page would take too long to load.

There are several ways you could filter that set and still have an infinite set. For instance, you could decide you just want numbers that are the product of three distinct primes. In a finite list of consecutive integers, there probably are fewer of those than there are squarefree numbers with at least two prime factors, but among all integers, both sets are infinite.

Robert Soupe
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