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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]

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    $\begingroup$ What is the function $I$? $\endgroup$ – Alexander Sep 9 '13 at 14:19
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    $\begingroup$ $I(n) = \frac{\sigma(n)}{n}$? $\endgroup$ – Daniel Fischer Sep 9 '13 at 14:21
  • $\begingroup$ 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. $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 9 '13 at 14:23
  • $\begingroup$ Hold on, I need to add an additional constraint. Again, my apologies. $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 9 '13 at 14:32
  • $\begingroup$ Done adding the constraint $I(n^2) < 2$. This last version of the question should be final. $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 9 '13 at 14:33
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This is extracted from my answer to Does this inequality hold true, in general?

Apply the result stated below to get the results for any constraints whatsoever.

I did a Google search for "density of euler phi function". The second link is http://www.ams.org/journals/proc/2007-135-09/S0002-9939-07-08771-0/S0002-9939-07-08771-0.pdf. This paper, by ANDREAS WEINGARTNER, is titled "THE DISTRIBUTION FUNCTIONS OF σ(n)/n AND n/ϕ(n)". Here is its abstract:

"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."

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