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A number is powerful if every prime divisor has multiplicity $\ge2$. I.e. if $n=\prod p_i^{k_i}$, then $n$ is powerful if all $k_i\ge2$.

I wanted to know if the the density of these numbers tends to $0, 1$, or something in-between? I found a lot of other info about them online but nothing about the density in particular.

Obviously the density of the perfect squares tends to $0$ but these powerful (or "squareful") numbers include more than just perfect squares. In fact, since every powerful number has a unique representation as $n=a^2 b^3$, it seems like the perfect squares are an infinitely small portion of the powerful numbers.

I also read that the sum of their reciprocals tends to a finite value: $$\sum_{a,b}\frac{1}{a^2b^3}=\prod_p(1-\frac{1}{p(p-1)}=\frac{315}{2\pi^4}\zeta(3)=1.943...$$ which I feel intuitively would indicate a density of zero since other sets, like the primes, have density zero but are frequent enough that the sum of their reciprocals diverges. I'm also not sure I understand the product above but guess it has something to do with powerful numbers being multaplicative with the squares and cubes of primes as primitive elements.

Chris Wolird
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    See this question. – lulu Sep 20 '22 at 17:03
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    To get a feeling how low the density of an infinite set can be with the sum of its reciprocals still diverging , consider this remarkable detail of Dirichlet's theorem : If $a$ and $b$ are coprime positive integers, there are not only infinite many primes of the form $an+b$ ($n$ positive integer), the sum of their reciprocals still diverges! This is even true for , say , $a$=Graham's number , $b=2$ !! – Peter Sep 20 '22 at 18:08
  • Remarkable is also that the sum of the reciprocals of the twin primes converges , considering the previous result extremely counterintuitive. – Peter Sep 20 '22 at 18:14
  • I'm really interested in the thought "[I]t seems like the perfect squares are an infinitely small portion of the powerful numbers." I feel like that might not be as likely as we think, though I'm not sure how to go about proving/disproving. For $n \le 5000$, there are $128$ powerful numbers, and over half ($70$) are perfect squares. Granted, the law of large numbers may start to take over at some point, but I wonder nonetheless. – Eric Snyder Sep 21 '22 at 10:39

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the powerful numbers are not $2 \pmod 4,$ single number crossed out, density $1 - \frac{1}{4}.$ Also not $3,6 \pmod 9,$ two crossed out, density $1 - \frac{2}{9}.$ Also not $5,10, 15, 20 \pmod {25},$ four crossed out, density $1 - \frac{4}{25}.$

The overall density is the product of $$ 1 - \frac{p-1}{p^2} = 1 + \frac{1-p}{p^2} $$

and density zero.

detail $ \log ( 1 - x) < -x $ for $0 < x < 1,$ log of the product is less than the sum of $ \; \; - \frac{p-1}{p^2} = \frac{1-p}{p^2}.$ In turn, this is less than $\frac{\pi^2}{6} $ minus the sum of $\frac{1}{p}$ and goes to $ \; - \infty $

Will Jagy
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