On the density of a particular subset of integers

Question

Given a positive integer $n$ in the standard form $$n=\prod_k p_k^{\alpha_k}$$ and the arithmetic function $$f(n)=\sum_k \alpha_k p_k$$ let's define the subset $F$ of positive integers $$F=\Big\{n\in N:f(n)\,|\,n,\;f(n)\lt n\Big\}=\Big\{16,27,30,60,70,72,84,105,150,\dots\Big\}$$ I ask if the density of this subset has ever been studied and, in particular, if it is possible to prove the convergence of the series $$\sum_{n\,\in\,F}\frac 1 n$$ Numerical experiments would show the convergence of such series towards a value quite close to the inverse of Euler's number $$\sum_{n\,\in\,F}\frac 1 n\sim\frac 1 e$$

Edit

My script is still running, but after $5\cdot 10^5$ terms ($n=584504910$) the sum of the series is $0.36652132586744884...\;(\frac 1 e = 0,36787944117144232...)$: the growth is extremely slow.

Work in progress

The most recent values obtained are the following:

$n=9928531324,\;\;3986000$-th term of the series$,\;\;$partial sum$\,=0.36776500537719703...$

$n=9931911561,\;\;3987000$-th term of the series$,\;\;$partial sum$\,=0.36776510608002266...$

$n=9935361024,\;\;3988000$-th term of the series$,\;\;$partial sum$\,=0.36776520674763440...$

$n=9938801814\,(\sim 10^{10}),\;\;3989000$-th term of the series$,\;\;$partial sum$\,=0.36776530738064540...$

I am cautiously optimistic about the convergence of the series.

Answer 1

Not an answer, but too long for a comment.

Given $m=\prod_{p} p^{a_p},$ we can find all $n=\prod_p p^{b_p}$ with $\sum_p pb_p=m$ and $m\mid n.$

Specifically, you want to solve for non-negative integers $c_p$ $$\sum_{p} c_pp=m-\sum_p a_pp$$ and then can use $b_p=c_p+a_p.$

So, for example, if $m=18,$ then you want to find solutions to:

$$\sum c_pp =18-2\cdot 3-2=10.$$

This means we can restrict to $p\in\{2,3,5,7\}$ and you get solutions:

$$(c_2,c_3,c_5,c_7)=(5,0,0,0), (2,2,0,0), (1,1,1,0), (0,1,0,1)$$

giving values:

$$n=2^6\cdot 3^2, 2^3\cdot 3^4, 2^2\cdot 3^3\cdot 5, 2\cdot 3^3\cdot 7.$$

Not sure how that helps. Clearly, there are no solutions when $m$ is prime. But even when $m=p_1p_2,$ there can be solutions. For example, $m=15$ has $n=3\cdot 5\cdot 7$ and $n=2\cdot 3\cdot 5^2$ as solutions.

You might be able to find some way to use groupings by $m$ to come up with some bounds on the sum.