1
$\begingroup$

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.]

Thanks!

$\endgroup$
2
$\begingroup$

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 $$\prod_1^k(p_i-\frac{1}{q_i})+\prod_1^k(p_iq_i-1)-\prod_1^k(p_iq_i-\frac{1}{q_i})\gt 0.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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