Bounds for two arithmetic functions, when one assumes that $n$ are odd perfect numbers

Question

For an integer $n>1$ in this post we denote the Dedekind psi function as $\psi(n)=n\prod_{\substack{p\mid n\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$ and the product of distinct primes dividing our integer $n$ as $\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p$, we've $\psi(1)=\operatorname{rad}(1)=1$. Both are important number theoretic functions in several subjects of mathematics. As reference I add the Wikipedia article Dedekind psi function (or [1]) and the corresponding Wikipedia Radical of an integer.

An integer $n\geq 1$ is said a perfect number if $\sigma(n)=\sum_{1\leq d\mid n}d=2n$. For example $n=6$ is the first perfect number and the only square-free (even) perfect number. It is unknown if odd perfect numbers do exist. Wikipedia has an article for Perfect number.

Using the properties of the aritmetic functions $\psi(n)$ and $\operatorname{rad}(n)$ one can to check easily the veracity of the following claim for even perfect numbers invoking the known as Euclid-Euler theorem.

Claim. A) If $n$ is an even perfect number then the identity $$\frac{\psi(n)}{n}=\frac{3}{2}+\frac{3(1+\sqrt{1+8n})}{8n}\tag{1}$$ holds.

B) If $n$ is an even perfect number then the identity $$\sum_{\substack{1\leq d\mid n\\d<\operatorname{rad}(n)}}\frac{1}{d}=2+\frac{3-\sqrt{1+8n}}{4n}\tag{2}$$ holds.

Previous formulas provides the best approximations of the arithmetic functions in the corresponding LHS of $(1)$ and $(2)$ for even perfect numbers because are identities (in terms of $n$ from the corresponding RHS).

Question. 1) What estimation/bounds can be done in terms of $n$ about $\frac{\psi(n)}{n}$ on assumption that $n$ is an odd perfect number? 2) And, what estimation/bounds can be done in terms pf $n$ about $\sum_{\substack{1\leq d\mid n\\d<\operatorname{rad}(n)}}\frac{1}{d}$ on assumption that $n$ is an odd perfect number? Many thanks.

I don't know if the problem to find bounds for the quantity $\prod_{\substack{p\mid n\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$, when one assumes that $n$ is an odd perfect number, is in the literature, in this case feel free to refer the literature answering the Question 1) as a reference request and I try to find and read those statements from the literature.

As motivation for those expressions I refer that the expression $(1)$ arises just from the quotient of the specialization of $\psi(n)$ for even perfect numbers by $n$, and on the other hand the sum in the LHS of $(2)$ evokes in some manner a variant of the sum $\sum_{1\leq d\mid n}\frac{1}{d}$ for perfect numbers or other more specific sums that are in the literature of odd perfect numbers.

References:

[1] Tom M. Apostol, Introduction to Analytic Number Theory, Undergraduate Texts in Mathematics Springer (1976).

Answer 1

As far as I'm aware, we don't have any substantially non-trivial bounds on the behavior of $\psi(n)$ when $n$ is an odd perfect number.

We can at least prove the following but none of these are difficult. Recall Euler's theorem that an odd perfect number must be of the form $N=q^em^2$ where $q$ is prime, $(q,m)=1$, and $q \equiv e \equiv 1$ (mod 4). Note that there's very little content to Euler's theorem here: it actually applies to any odd number $n$ where $\sigma(n) \equiv 2$ (mod 4). In what follows, we'll write $N$ as an odd perfect number, and reserve $q$ and $e$ and $m$ as the numbers associated to $N$ as defined above. Note that $q$ is sometimes referred to as the "Euler prime" of an odd perfect number and sometimes as the "special prime." In fact, before Euler's result, Descartes proved the weaker result that an odd perfect number must have exactly one prime factor raised to an odd power, so a better name might be the Cartesian prime, but this term seems to be unpopular.

It is not hard to see that $\psi(N)$ and $N$ must share many prime factors. One easily has that for example $(q+1)m|\psi(N)$, and it isn't hard to see that $\frac{q+1}{2}|N$ (although note one cannot conclude that $\frac{q+1}{2}m|N$ since $q+1$ may have a squared prime factor).

In general, obviously $\psi(n) \leq \sigma(n)$ with equality iff $n$ is prime or $n$ is one. We'll write $t(n) = \frac{\sigma(n)}{\psi(n)}$. Note that if $p$ is an odd prime prime and $k >1$ then we have $$t(p^k) = \frac{p^k + p^{k-1} + \cdots + p + 1}{p^k + p^{k-1}} = 1 + \frac{p^{k-2} + \cdots 1}{p^k + p^{k-1}} \leq 1+ \frac{1}{p^2}.$$

Since $\prod_{p} 1+\frac{1}{p^2}$ converges, one never has for any $n$ (perfect or not) $\psi(n)$ much larger than $\sigma(n)$. One can show for an odd perfect number that they must be somewhat closer to each other than one would get just from this sort of argument. For example, using the fact that an odd perfect number cannot be divisible by all of $3$, $5$ and $7$, one never has all three of those primes in the actually relevant product.

But none of this content is really non-trivial. This is about what one would expect an undergrad who has taken a basic number theory course to prove if they became interested in the subject.