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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.

(A similar question is posted in MSE here.)

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    $\begingroup$ What is the "abundancy index"? In particular, what is $\sigma(x)$? And why did you post on MO within 7 minutes of posting on MSE instead of waiting for responses there? $\endgroup$
    – Vidit Nanda
    Commented Sep 9, 2013 at 15:39
  • $\begingroup$ It seems likely that the asymptotic density is 0 using standard arguments. What have you tried? $\endgroup$
    – The Masked Avenger
    Commented Sep 9, 2013 at 16:39
  • $\begingroup$ @ViditNanda, my short answer to your last question is that I am not so sure if this question can be considered a "research-level" question as per MO standards. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Sep 9, 2013 at 16:57
  • $\begingroup$ @TheMaskedAvenger, I've tried polynomial division on the middle inequality, thereby getting the final result $$2n^2 - 2n + 1 \leq \sigma(n^2) < 2n^2.$$ This means that the abundance $a(n^2) = \sigma(n^2) - 2n^2$ satisfies $1 - 2n \leq a(n^2) < 0$, while the deficiency $d(n^2) = 2n^2 - \sigma(n^2)$ satisfies $0 < d(n^2) \leq 2n - 1$. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Sep 9, 2013 at 17:03
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    $\begingroup$ @StevenLandsburg : Thank you for pointing that out, I certainly did not mean to be rude. I am just another amateur MO/MSE OP looking for an answer to my question. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Sep 10, 2013 at 4:18

1 Answer 1

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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 $0$.

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  • $\begingroup$ Thank you for your answer @so-called-friend-don, appreciate it! =) $\endgroup$
    – Jose Arnaldo Bebita
    Commented Sep 10, 2013 at 4:19
  • $\begingroup$ I forgot to ask (and my apologies in advance if this is too dumb a question), but does your result imply that the asymptotic density or limiting proportion of $n$ satisfying the inequality $$I(n^2) < \frac{2n}{n + 1},$$ is $1$? $\endgroup$
    – Jose Arnaldo Bebita
    Commented Sep 17, 2013 at 10:03

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