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](https://math.stackexchange.com/questions/490720).]


Thanks!



What proportion of the positive integers satisfy $I(n^2) < (1 + \frac{1}{n})I(n)$, where $I(x)$ is the abundancy index of $x$?
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 $\frac{1}{\ln N}$




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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 $$\prod\_1^k(p\_i-\frac{1}{q\_i})+\prod\_1^k(p\_iq\_i-1)-\prod\_1^k(p\_iq\_i-\frac{1}{q\_i})\gt 0.$$ 


