# What proportion of the positive integers satisfy $I(n) < \frac{2n}{n + 1} \leq I(n^2)$ < 2?

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.

[Edit: September 10 2013] This question has already been answered by Don in MO here. Thanks everyone! [End edit]

• What is the function $I$? – Alexander Sep 9 '13 at 14:19
• $I(n) = \frac{\sigma(n)}{n}$? – Daniel Fischer Sep 9 '13 at 14:21
• My apologies for missing out on the definition of the function $I$. Yes @DanielFischer, indeed it is the abundancy index function. Editing my question to reflect that change now. – Jose Arnaldo Bebita-Dris Sep 9 '13 at 14:23
• Hold on, I need to add an additional constraint. Again, my apologies. – Jose Arnaldo Bebita-Dris Sep 9 '13 at 14:32
• Done adding the constraint $I(n^2) < 2$. This last version of the question should be final. – Jose Arnaldo Bebita-Dris Sep 9 '13 at 14:33

"Let σ(n) be the sum of the positive divisors of n. We show that the natural density of the set of integers n satisfying σ(n)/n ≥ t is given by $\exp\big(−e^{t e^{−γ}(1 + O(t^{−2}))}\big)$ , where γ denotes Euler’s constant. The same result holds when σ(n)/n is replaced by n/ϕ(n), where ϕ is Euler’s totient function."