Let $\sigma(x)$ denote the classical sum-of-divisors function, and let

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

be the abundancy index of the positive integer $x$.

My question is this: What proportion of the positive integers satisfy

$$I(n^2) < (1 + \frac{1}{n})I(n),$$

if, in addition, we know that both $n$ and $n^2$ are deficient numbers?

Note that, trivially, we have $I(n) \leq I(n^2) \leq (I(n))^2$ for all integers $n \geq 1$.

[This question was cross-posted from MSE.]



I think that you will find that you are looking at the primes and powers of primes. Their density in the integers up to $N$ is essentially the same as the density of primes, roughly $\frac{1}{\ln N}$

For a prime power $q=p^e$ (including the case $q=p^1$) we have $$\sigma(q)=\frac{pq-1}{p-1}.$$ For $n=\prod_1^kq_i$ a product of powers of distinct primes we have $\sigma(n)=\prod_1^k\frac{p_iq_i-1}{p_i-1}$ and $\sigma(n^2)=\prod_1^k\frac{p_iq_i^2-1}{p_i-1}.$

Your condition is equivalent to $$(1+n)\sigma(n)-\sigma(n^2) \gt 0$$ which becomes $$\left(1+\prod_1^kq_i\right)\prod_1^k(p_iq_i-1)-\prod_1^k(p_iq_i^2-1) \gt 0$$

That is true for $k=1$. I'm not quite sure the best way to show that it fails for $k \gt 1$, but it seems clear that it does, It might help to manipulate it to $$\prod_1^k(p_iq_i-1)+\prod_1^k(p_iq_i^2-q_i)-\prod_1^k(p_iq_i^2-1)\gt 0$$ and perhaps then to $$\prod_1^k(p_i-\frac{1}{q_i})+\prod_1^k(p_iq_i-1)-\prod_1^k(p_iq_i-\frac{1}{q_i})\gt 0.$$

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  • $\begingroup$ Just a follow-up question on your answer, @AaronMeyerowitz - does this mean that, $\omega(n) \geq 2$ if and only if $$\left(1 + \frac{1}{n}\right)I(n) \leq I(n^2),$$ where $\omega(y)$ is the number of distinct prime factors of $y$ and $I(x) = \frac{\sigma(x)}{x}$ is the abundancy index of $x$? $\endgroup$ – Arnie Bebita-Dris Oct 2 '13 at 12:18
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    $\begingroup$ Yes, it would indeed mean that. $\endgroup$ – Aaron Meyerowitz Oct 3 '13 at 2:40
  • $\begingroup$ My profuse thanks, @AaronMeyerowitz! =) $\endgroup$ – Arnie Bebita-Dris Oct 3 '13 at 4:37

Let $d$ be the smallest divisor of $n^2$, which does not divide $n$. Then $I(n^2)>I(n)+\frac{1}{d}$, hence, if $I(n^2)\leq(1+\frac{1}{n})I(n)$, then $d<\frac{I(n)}{n}$.But $I(n)=\mathcal{O}(\log\log n)$, hence $d$ must be surprisingly large. In particular, if $n$ has three different prime factors, then $d<n^{2/3}$, which gives a contradiction for $n$ sufficiently large. If $n=p^aq^b$,then $d=\min(p^{a+1}, q^{b+1})$,and $I(n)\leq 3$, thus $a=b=1$ and $p,q\approx\sqrt{n}$.If $n=p^a$, then $$ \frac{I(n^2)}{I(n)}=\frac{p-p^{-2a}}{p-p^{-a}}= 1+\frac{p^{-a}-p^{-2a}}{p-p^{-a}}<1+p^{-a}. $$ Hence the integers in question are prime powers and certain product of two primes of similar size. The precise determination of "similar" should be possible by a straightforward yet lengthy computation.

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    $\begingroup$ Primes and prime powers do have this property. But a product of two primes never does. Rewrite the required inequality as $(n+1)\sigma(n)-\sigma(n^2) \gt 0.$ For $n=ab$ a product of two primes (similar in size or not) this becomes $(ab+1)(a+1)(b+1)-(a^2+a+1)(b^2+b+1) \gt 0$ I.E. $ab-a^2-b^2\gt 0$ $\endgroup$ – Aaron Meyerowitz Oct 2 '13 at 8:49
  • $\begingroup$ Thank you very much for your answer, @Jan-ChristophSchlage-Pu! $\endgroup$ – Arnie Bebita-Dris Oct 2 '13 at 17:23

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