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I apologize in advance if this is something that is already well-known in the literature, but I would like to ask nonetheless (for the benefit of those who likewise do not know):

Are there (known) lower and upper bounds to the following arithmetic / number-theoretic expression:

$$\frac{I(x^2)}{I(x)} = \frac{\frac{\sigma_1(x^2)}{x^2}}{\frac{\sigma_1(x)}{x}}$$

where $x \in \mathbb{N}$, $\sigma_1(x)$ is the sum of the divisors of $x$ and $I(x) = \frac{\sigma_1(x)}{x}$ is the abundancy index of $x$?

(Note that a trivial lower bound is $1$ since $x \mid x^2$ implies $I(x) \leq I(x^2)$.)

I would highly appreciate it if somebody will be able to point me to relevant references in the existing literature.

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I just realized that $$\displaystyle\frac{I(x^2)}{I(x)} < I(x).$$ – user11235813 Mar 5 '13 at 14:23
Consequently, a conclusive answer to this question would depend on the Riemann Hypothesis (RH), per Robin's original formulation of a condition involving the sum-of-divisors function, and which Robin showed to be equivalent to RH. – user11235813 Mar 5 '13 at 14:26
There's no need to apologize for asking a question you don't know the answer to! – Qiaochu Yuan Feb 8 '15 at 2:32
up vote 1 down vote accepted

Will Jagy has posted an answer to a closely related MSE question here.

Per Will Jagy's answer in the linked MSE question (and a subsequent comment by Erick Wong), we have the conjectured (sharp?) bounds

$$1 \leq \frac{I(x^2)}{I(x)} \leq \prod_{p}{\frac{p^2 + p + 1}{p^2 + p}} = \frac{\zeta(2)}{\zeta(3)} \approx 1.3684327776\ldots$$

I would still be interested in an (improved) lower bound for $I(x^2)/I(x)$ when $\omega(x) \geq 3$, where $\omega(x)$ is the number of distinct prime factors of $x$.

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