Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.)

share|improve this question
2  
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? –  Vidit Nanda Sep 9 '13 at 15:39
    
It seems likely that the asymptotic density is 0 using standard arguments. What have you tried? –  The Masked Avenger Sep 9 '13 at 16:39
    
@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. –  Jose Arnaldo Dris Sep 9 '13 at 16:57
    
@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 $0 < d(n^2) \leq 2n - 1$. –  Jose Arnaldo Dris Sep 9 '13 at 17:03
1  
@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. –  Jose Arnaldo Dris Sep 10 '13 at 4:18

1 Answer 1

up vote 4 down vote accepted

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 $0$.

share|improve this answer
    
Thank you for your answer @so-called-friend-don, appreciate it! =) –  Jose Arnaldo Dris Sep 10 '13 at 4:19
    
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 $1$? –  Jose Arnaldo Dris Sep 17 '13 at 10:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.