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.
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 $