Let


$$I(x) = \frac{\sigma(x)}{x}$$


be the abundancy index of the positive integer $x$. Note that $\sigma(x)$ is the classical sum-of-divisors function. For example,


$$\sigma(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28.$$


My question is this: What proportion of the positive integers satisfy the inequality
$$I(n) < \frac{2n}{n + 1} \leq I(n^2) < 2?$$


Note that we necessarily have $n > 1$ from the left-hand inequality.


(A similar question is posted in MSE [here](https://math.stackexchange.com/questions/488443).)



What proportion of the positive integers satisfy $I(n) < \frac{2n}{n + 1} \leq I(n^2) < 2$?
The function $I(n^2)$ has a continuous limiting distribution (look up the Erd\H{o}s--Wintner theorem). Since your lower limit $2n/(n+1)$ converges to your upper limit of $2$ as $n\to\infty$, the continuity of the distribution function shows that the limiting proportion of $n$ satisfying your inequality is $0$.


