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In this post, for integers $n\geq 1$, I denote the sum of divisors $\sum_{1\leq d\mid n}d$ as $\sigma(n)$ and the Euler's totient function as $\varphi(n)$. It's easy to check* that if we assume that $n=N$ is an odd perfect number then the inequalities

$$\frac{\sigma\left(\frac{\sigma(n)}{2}n^2\right) }{\frac{\sigma(n)}{2}(\sigma(n)-n)^2}\leq \frac{n}{\varphi(\sigma(n)-n)}\tag{1}$$

and

$$8\frac{\sigma\left(\frac{\sigma(n)}{2}n^2\right) }{\frac{\sigma(n)}{2}(\sigma(\sigma(n))-\sigma(n))^2}\leq \frac{n}{\varphi(\sigma(\sigma(n))-\sigma(n))},\tag{2}$$ hold. And it is obvious that every odd perfect number $n=N$ is an odd integer satisfying

$$2\mid \sigma(n).\tag{3}$$

Example. The odd integer $n=1102725$ is the least positive integer satisfying $(1)$ and $(3)$. But (I know that) this odd integer doesn't satisfy $(2)$.

I'm asking what work can be done for the following questions (the part A) is just a computational exercise) before I'm accepting an answer.

Question. A) Please, I would like to know if you can to calculate the least odd integer $n\geq 1$ satisfying the inequality $(1)$, the condition $(3)$ and also the inequality $(3)$. B) I would like to know if it is possible to elucidate (what work can be done about this) if the set of solutions $n$ satisfying all requirements $n\equiv 1\text{ mod }2$, the inequalities $(1)$, $(2)$ and $2\mid \sigma(n)$) is bounded, that's: are there finitely/infinitely many solutions over odd integers $n\geq 1$ satisfying both inequalities together the cited condition $(3)$? Many thanks.

The motivation of A) is that I'm not able (with my computer and knowledges) to calculate it.

*The calculation is easy from a well known inequality for the sum of divisors function and the Euler's totient (Exercise 9 for Chapter 3 of [1]), since for perfect numbers $\sigma(N)/2=\sigma(N)-N=N$ and thus our inequalities are simplified by using $x^a/\varphi(x^a)=x/\varphi(x)$.

Remark: I edit this (if can be inspiring here), to add that it is possible to state different formulae involving primorials. I hope don't bother with these aside remarks.

1. It is easy to check that if $n=N$ is an odd perfect number that satisfies $\left(n,\frac{N_l}{N_k}\right)=1$ for some integers $1\leq k<l$ with $N_t=\prod_{j=1}^t p_j$ denoting the primorial of order $t$ then the following identity holds $$2\frac{\varphi(N_k)}{\varphi(N_l)}\varphi\left(2n^2\frac{N_l}{N_k}\right)=\sigma(n)\varphi(n).$$

2. From this identity we claim that it is easy to check the following inequality for odd perfect numbers $n=N$ under previous assumption $\left(n,\frac{N_l}{N_k}\right)=1$ $$\frac{\sigma\left(n^2\frac{\sigma(n)}{2}\right)}{\left(\sigma(n)-n\right)^3}\leq\frac{n\sigma(n)}{2\frac{\varphi(N_k)}{\varphi(N_l)}\varphi\left(2(\sigma(n)-n)^2\frac{N_l}{N_k}\right)}.$$

References:

[1] Tom M. Apostol, Introduction to Analytic Number Theory, UTM, Springer (1976).

In this post, for integers $n\geq 1$, I denote the sum of divisors $\sum_{1\leq d\mid n}d$ as $\sigma(n)$ and the Euler's totient function as $\varphi(n)$. It's easy to check* that if we assume that $n=N$ is an odd perfect number then the inequalities

$$\frac{\sigma\left(\frac{\sigma(n)}{2}n^2\right) }{\frac{\sigma(n)}{2}(\sigma(n)-n)^2}\leq \frac{n}{\varphi(\sigma(n)-n)}\tag{1}$$

and

$$8\frac{\sigma\left(\frac{\sigma(n)}{2}n^2\right) }{\frac{\sigma(n)}{2}(\sigma(\sigma(n))-\sigma(n))^2}\leq \frac{n}{\varphi(\sigma(\sigma(n))-\sigma(n))},\tag{2}$$ hold. And it is obvious that every odd perfect number $n=N$ is an odd integer satisfying

$$2\mid \sigma(n).\tag{3}$$

Example. The odd integer $n=1102725$ is the least positive integer satisfying $(1)$ and $(3)$. But (I know that) this odd integer doesn't satisfy $(2)$.

I'm asking what work can be done for the following questions (the part A) is just a computational exercise) before I'm accepting an answer.

Question. A) Please, I would like to know if you can to calculate the least odd integer $n\geq 1$ satisfying the inequality $(1)$, the condition $(3)$ and also the inequality $(2)$. B) I would like to know if it is possible to elucidate (what work can be done about this) if the set of solutions $n$ satisfying all requirements $n\equiv 1\text{ mod }2$, the inequalities $(1)$, $(2)$ and $2\mid \sigma(n)$) is bounded, that's: are there finitely/infinitely many solutions over odd integers $n\geq 1$ satisfying both inequalities together the cited condition $(3)$? Many thanks.

The motivation of A) is that I'm not able (with my computer and knowledges) to calculate it.

*The calculation is easy from a well known inequality for the sum of divisors function and the Euler's totient (Exercise 9 for Chapter 3 of [1]), since for perfect numbers $\sigma(N)/2=\sigma(N)-N=N$ and thus our inequalities are simplified by using $x^a/\varphi(x^a)=x/\varphi(x)$.

Remark: I edit this (if can be inspiring here), to add that it is possible to state different formulae involving primorials. I hope don't bother with these aside remarks.

1. It is easy to check that if $n=N$ is an odd perfect number that satisfies $\left(n,\frac{N_l}{N_k}\right)=1$ for some integers $1\leq k<l$ with $N_t=\prod_{j=1}^t p_j$ denoting the primorial of order $t$ then the following identity holds $$2\frac{\varphi(N_k)}{\varphi(N_l)}\varphi\left(2n^2\frac{N_l}{N_k}\right)=\sigma(n)\varphi(n).$$

2. From this identity we claim that it is easy to check the following inequality for odd perfect numbers $n=N$ under previous assumption $\left(n,\frac{N_l}{N_k}\right)=1$ $$\frac{\sigma\left(n^2\frac{\sigma(n)}{2}\right)}{\left(\sigma(n)-n\right)^3}\leq\frac{n\sigma(n)}{2\frac{\varphi(N_k)}{\varphi(N_l)}\varphi\left(2(\sigma(n)-n)^2\frac{N_l}{N_k}\right)}.$$

References:

[1] Tom M. Apostol, Introduction to Analytic Number Theory, UTM, Springer (1976).

In this post, for integers $n\geq 1$, I denote the sum of divisors $\sum_{1\leq d\mid n}d$ as $\sigma(n)$ and the Euler's totient function as $\varphi(n)$. It's easy to check* that if we assume that $n=N$ is an odd perfect number then the inequalities

$$\frac{\sigma\left(\frac{\sigma(n)}{2}n^2\right) }{\frac{\sigma(n)}{2}(\sigma(n)-n)^2}\leq \frac{n}{\varphi(\sigma(n)-n)}\tag{1}$$

and

$$8\frac{\sigma\left(\frac{\sigma(n)}{2}n^2\right) }{\frac{\sigma(n)}{2}(\sigma(\sigma(n))-\sigma(n))^2}\leq \frac{n}{\varphi(\sigma(\sigma(n))-\sigma(n))},\tag{2}$$ hold. And it is obvious that every odd perfect number $n=N$ is an odd integer satisfying

$$2\mid \sigma(n).\tag{3}$$

Example. The odd integer $n=1102725$ is the least positive integer satisfying $(1)$ and $(3)$. But (I know that) this odd integer doesn't satisfy $(2)$.

I'm asking what work can be done for the following questions (the part A) is just a computational exercise) before I'm accepting an answer.

Question. A) Please, I would like to know if you can to calculate the least odd integer $n\geq 1$ satisfying the inequality $(1)$, the condition $(3)$ and also the inequality $(3)$. B) I would like to know if it is possible to elucidate (what work can be done about this) if the set of solutions $n$ satisfying all requirements $n\equiv 1\text{ mod }2$, the inequalities $(1)$, $(2)$ and $2\mid \sigma(n)$) is bounded, that's: are there finitely/infinitely many solutions over odd integers $n\geq 1$ satisfying both inequalities together the cited condition $(3)$? Many thanks.

The motivation of A) is that I'm not able (with my computer and knowledges) to calculate it.

*The calculation is easy from a well known inequality for the sum of divisors function and the Euler's totient (Exercise 9 for Chapter 3 of [1]), since for perfect numbers $\sigma(N)/2=\sigma(N)-N=N$ and thus our inequalities are simplified by using $x^a/\varphi(x^a)=x/\varphi(x)$.

References:

[1] Tom M. Apostol, Introduction to Analytic Number Theory, UTM, Springer (1976).

In this post, for integers $n\geq 1$, I denote the sum of divisors $\sum_{1\leq d\mid n}d$ as $\sigma(n)$ and the Euler's totient function as $\varphi(n)$. It's easy to check* that if we assume that $n=N$ is an odd perfect number then the inequalities

$$\frac{\sigma\left(\frac{\sigma(n)}{2}n^2\right) }{\frac{\sigma(n)}{2}(\sigma(n)-n)^2}\leq \frac{n}{\varphi(\sigma(n)-n)}\tag{1}$$

and

$$8\frac{\sigma\left(\frac{\sigma(n)}{2}n^2\right) }{\frac{\sigma(n)}{2}(\sigma(\sigma(n))-\sigma(n))^2}\leq \frac{n}{\varphi(\sigma(\sigma(n))-\sigma(n))},\tag{2}$$ hold. And it is obvious that every odd perfect number $n=N$ is an odd integer satisfying

$$2\mid \sigma(n).\tag{3}$$

Example. The odd integer $n=1102725$ is the least positive integer satisfying $(1)$ and $(3)$. But (I know that) this odd integer doesn't satisfy $(2)$.

I'm asking what work can be done for the following questions (the part A) is just a computational exercise) before I'm accepting an answer.

Question. A) Please, I would like to know if you can to calculate the least odd integer $n\geq 1$ satisfying the inequality $(1)$, the condition $(3)$ and also the inequality $(3)$. B) I would like to know if it is possible to elucidate (what work can be done about this) if the set of solutions $n$ satisfying all requirements $n\equiv 1\text{ mod }2$, the inequalities $(1)$, $(2)$ and $2\mid \sigma(n)$) is bounded, that's: are there finitely/infinitely many solutions over odd integers $n\geq 1$ satisfying both inequalities together the cited condition $(3)$? Many thanks.

The motivation of A) is that I'm not able (with my computer and knowledges) to calculate it.

*The calculation is easy from a well known inequality for the sum of divisors function and the Euler's totient (Exercise 9 for Chapter 3 of [1]), since for perfect numbers $\sigma(N)/2=\sigma(N)-N=N$ and thus our inequalities are simplified by using $x^a/\varphi(x^a)=x/\varphi(x)$.

Remark: I edit this (if can be inspiring here), to add that it is possible to state different formulae involving primorials. I hope don't bother with these aside remarks.

1. It is easy to check that if $n=N$ is an odd perfect number that satisfies $\left(n,\frac{N_l}{N_k}\right)=1$ for some integers $1\leq k<l$ with $N_t=\prod_{j=1}^t p_j$ denoting the primorial of order $t$ then the following identity holds $$2\frac{\varphi(N_k)}{\varphi(N_l)}\varphi\left(2n^2\frac{N_l}{N_k}\right)=\sigma(n)\varphi(n).$$

2. From this identity we claim that it is easy to check the following inequality for odd perfect numbers $n=N$ under previous assumption $\left(n,\frac{N_l}{N_k}\right)=1$ $$\frac{\sigma\left(n^2\frac{\sigma(n)}{2}\right)}{\left(\sigma(n)-n\right)^3}\leq\frac{n\sigma(n)}{2\frac{\varphi(N_k)}{\varphi(N_l)}\varphi\left(2(\sigma(n)-n)^2\frac{N_l}{N_k}\right)}.$$

References:

[1] Tom M. Apostol, Introduction to Analytic Number Theory, UTM, Springer (1976).