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As tell us the Wikipedia section dedicated to Odd perfect numbers (please, see also the related references if you need it), any perfect number has the form $$n=q^\alpha m^2$$ where the integer $\alpha\geq 1$ satisfies $\alpha\equiv 1\text{ mod }4$, the integer $q$ is a prime number satisfying $q\equiv 1\text{ mod }4$ and the positive integer $m$ satisfies $\gcd(q,m)=1$. This is the Euler's theorem for odd perfect numbers.

The factor $q^\alpha$, in the factorization of Euler's theorem for odd perfect numbers, is the so-called Euler factor of our odd perfect number $n$.

In this paragraph we remember the definitions/notations for some number theoretic function: for real numbers $x\geq 1$, in this post the prime-counting function is denoted as $\pi(x)$. For an integer $n>1$, we denote its greatest prime factor as $\operatorname{gpf}(n)$, and the product of the distinct prime numbers dividing $n$ as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ with the definition $\operatorname{rad}(1)=1$ (see it you want the Wikipedia Radical of an integer and the corresponding article from MathWorld Greatest Prime Factor).

Question. Is it possible, under the assumption that $n$ is an odd perfect number, to get bounds for the inequality $$\text{lower bound}<\pi\left(q^\alpha\right)<\text{upper bound},\tag{1}$$ being $q^{\alpha}$ its Euler factor, or well for inequalities of the type $$\text{lower bound}<\pi\left(\operatorname{rad}(n)\right)<\text{upper bound},\tag{2}$$ or $$\text{lower bound}<\pi\left(\operatorname{gpf}(n)\right)<\text{upper bound}\,?\tag{3}$$ Many thanks.

The calculations that I evoke are deductions of the bounds as functions of $n$ (including constants, if it is the case). My motivation is to learn how do it, and that if there is some answer it can be a very good reference of your proposition/s for other people interested in this theory, I think.

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  • $\begingroup$ Of course, the expressions of such bounds also can be written in terms of the factorization of our odd perfect numbers, I am saying that you can express your bounds invonving the primes of the primes and their corresponding exponents of the prime factorization of $n$. Many thanks again y thanks for the patience of all users. $\endgroup$ – user142929 Aug 10 at 9:16
  • $\begingroup$ Many thanks for your edit @MartinSleziak $\endgroup$ – user142929 Sep 10 at 16:24
  • $\begingroup$ Also, the radical of a number and the notion of greatest prime factor are well-known enough ideas that you don't really need to reference them, (and definitely don't need to reference Wikipedia). $\endgroup$ – JoshuaZ Sep 11 at 1:05
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Since we have good asymptotics for $\pi(n)$ by the prime number theorem (and can get good explicit bounds on that from Rosser and Schoenfeld's work as well as later work such as that by Dusart) this question is essentially the same as asking for interesting upper and lower bounds on $q^a$, $\mathrm{rad}(n)$ and $\mathrm{gpf}(n)$. For convenience, I'm going to write $R=\mathrm{rad}(n)$ and $P=\mathrm{gpf}(n)$

In the case of $q^a$ there is no general lower bound known. At Tthe current state of the literature, we cannot even rule out the existence of an odd perfect number where $q^a=5$, the smallest candidate.

There's been some work on upper and lower bounding $R$. There's a paper by Luca and Pomerance (On the radical of a perfect number, New York Journal of Mathematics 16 (2010) 23–30. ) which shows that $R < 2n^{\frac{17}{26}}.$ This was subsequently improved in part by a paper by Klurman (O. Klurman, Radical of perfect numbers and perfect numbers among polynomial values, International Journal of Number Theory 12 3 (2016), 585–591), which replaces the $\frac{17}{26}$ with $\frac{9}{14}$ but with the cost of having a non-explicit constant in front; Klurman's argument can be made explicitly with a little work, and it isn't hard to tighten up his result to get that $R < 4n^{\frac{9}{14}}.$ For the last few years there's been a goal of a variety of authors to get a result of the form $R < Cn^{\frac{1}{2}}.$ This would be a natural result because one can translate results bounding $P < n^{\epsilon}$ into results of the form $R < C_{\epsilon}n^{\frac{1}{2}+ \epsilon}.$ See both Klurman's paper as well a related paper by Acquaa and Konyagin ( On prime factors of odd perfect numbers, International Journal of Number Theory 8 6 (2012), 1537–1540.)

Lower bounding $R$ is also in the literature although what sort of lower bound you want matters a bit. If you want a lower bound in terms of $n$ itself, then the most natural one is due to Pace Nielsen, who has shown that if $n$ has $k$ distinct prime factors, then $$n < R^{2^k-1}.$$ Pace's bound can be naturally turned into a lower bound on $R$ since one has an upper bound for $k$ in terms of $R$. Another, less direct way to lower bound $R$ is directly in terms of $k$. In this case, I don't know of any non-trivial bound.

Lastly for $P$, the only non-trivial bound in the literature is in the earlier mentioned bound of Acquaa and Konyagin who proved that $P < (3N)^{1/3}$. Lower bounding $P$ can be done in terms of $k$, and one can then use lower bounds for $k$ in terms of $N$ (see also work by Pace Nielsen). There's a paper of mine currently under review which gets an improved lower bound on $P$ in terms of the smallest prime factor of $N$, although that aspect is really just tightening up some earlier results of Norton from the 1960s since we have tighter bounds on the PNT since Norton's paper. The preprint for that is here (The other results there are likely of more interest).

I should also note that there's been some work with the second largest and third largest prime factor of an odd perfect number. I proved a non-trivial upper bound on the second largest prime factor which you can find here. There's another paper currently under review by me, Sean Bibby and Pieter Vyncke which upper bounds the third largest prime factor. Preprint is here. Note that all three of these papers of I consider to have lots of questions of general interests to number theorists even if they aren't interested in odd perfect numbers, so if anyone in that set is still reading this far down, by all means check them out.

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  • $\begingroup$ Many thanks for your answer, I'm going to study it in next days. $\endgroup$ – user142929 Sep 11 at 9:14

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