# Density of Numbers with Exactly One Prime Factor of Multiplicity 1

Let $$S$$ be the set of positive integers $$n$$ with the property that exactly one prime factor of $$n$$ has multiplicity $$1$$ and every other prime factor has multiplicity greater than $$1$$ (to be clear, $$S$$ does not contain the prime numbers). What is the asymptotic behavior of $$\frac{\#\{n \in S' : n \leq X\}}{\#\{n \leq X\}}$$ in the limit as $$X \to \infty$$?

What I know: There is an easier analogue of this problem that I have an approach for, namely finding the density of the set $$S'$$ of positive integers $$n$$ such that every prime factor of $$n$$ has multiplicity greater than $$1$$. It is not hard to check that $$\frac{\#\{n \in S' : n \leq X\}}{\#\{n \leq X\}} \leq \int_{1}^X (X/n^2)^{1/3} dn \approx 3 \sqrt{X}.$$ I suspect that the density of $$S$$ must be in the literature somewhere, but I can at least use the density of $$S'$$ to get an upper bound on the density of $$S$$ as follows: $$\frac{\#\{n \in S : n \leq X\}}{\#\{n \leq X\}} \leq \sum_{p \text{ prime}} \frac{\#\{n \in S' : n \leq X/p\}}{\#\{n \leq X/p\}} \leq \sum_{p \text{ prime}}3 \sqrt{X/p} \ll X/\log X,$$ where in the last step above, I've used a well-known estimate for the sum of the reciprocals of the square roots of the prime numbers. Can one do better than $$X/\log X$$? (Is a power savings possible?)

• aka square free multipliers on squareful numbers. – Roddy MacPhee Jul 3 at 23:24
• OEIS A056169 suggests that up to $X=10000$ you have $2956$ such numbers if you include the primes and so $1727$ if you exclude the primes. This is rather larger than $\frac{10000}{\log_e 10000} \approx 1085.7$. Or are you suggesting $0.1727 \ll 1085.7$, which is certainly true though not enlightening? – Henry Jul 3 at 23:25
• @Henry Thanks for the OEIS reference!! The $\ll$ sign includes an implied constant that's approximately equal to $2$, so there isn't a contradiction. – Ashvin Swaminathan Jul 3 at 23:29
• Further savings is definitely not possible, since this set contains every integer of the form $4p$ where $p$ is an odd prime, and so the counting function is $\ge \pi(x/4)-1 \gg x/\log x$. By the way, your set $S'$ is the set of powerful numbers, and its counting function is known to be asymptotic to $\frac{\zeta(3/2)}{\zeta(3)}\sqrt x$. – Greg Martin Jul 4 at 0:41
• Let $F(s) =1+ \sum_{n \in S'}n^{-s}= \prod_p (1+\frac{p^{-2s}}{1-p^{-s}}), \quad P(s) = \sum_p p^{-s}$ you want to look at $$\sum_{n \in S}n^{-s}= \sum_p p^{-s}( \frac{F(s)}{1+\frac{p^{-2s}}{1-p^{-s}}}-1) = -P(s) + F(s) P(s)+F(s)R(s)$$ where $R(s) = \sum_p p^{-s} (\frac{1}{1+\frac{p^{-2s}}{1-p^{-s}}}-1)$ is a Dirichlet series converging absolutely for $\Re(s) > 1/3$. With the standard tauberian theorems it gives $\sum_{n \le x,n\in S} 1 \sim (F(1)-1)\frac{x}{\log x}$ – reuns Jul 4 at 1:32