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0 answers
25 views

Improving $I(m^2)/I(m) < 2^{\log(13/12)/\log(13/9)}$ where $p^k m^2$ is an odd perfect number with special prime $p$

In what follows, let $I(x)=\sigma(x)/x$ denote the abundancy index of the positive integer $x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$. The following is an attempt to ...
Jose Arnaldo Bebita Dris's user avatar
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0 answers
51 views

Divisors of $x^2-1$ in Brocard's Problem

In this post, I was curious if the divisor bound could be improved for the product of two consecutive even numbers. It seems most likely it cannot by much. How could the upper bound of $\sigma_0(x^2 - ...
PiMaster's user avatar
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0 answers
35 views

Maxima of the Sum of Divisors Function and Upper Bound of a similar Ratio

If we define a function (aka Gronwall’s function) as: $$F(n)=\frac{\sigma(n)}{n \log \log n}$$ Then for $n>15$, it does have an upper bound. I want to know what's that specific upper bound is? Also ...
Ok-Virus2237's user avatar
0 votes
2 answers
102 views

Is there an analytical solution to the inequality $\frac{1 + k({p}^{(k+1)/2)})}{p^k} < \frac{p}{p - 1}$ if one were to bound $k$ in terms of $p$?

My question is as is in the title: Is there an analytical solution to the inequality $$\frac{1 + k({p}^{(k+1)/2)})}{p^k} < \frac{p}{p - 1}$$ if one were to bound $k$ in terms of $p$? Here, $p \...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
223 views

Can this inequality involving the deficiency and sum of aliquot divisors be improved? - Part II

This MSE question (from April 2020) asked whether the inequality $$\frac{D(n^2)}{s(n^2)} < \frac{D(n)}{s(n)}$$ could be improved, where $D(x)=2x-\sigma(x)$ is the deficiency of the positive integer ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
0 answers
119 views

Does the following lower bound improve on $I(q^k)+I(n^2) > \frac{57}{20}$, where $q^k n^2$ is an odd perfect number?

Preamble: This question is an offshoot of this earlier post. (This inquiry has likewise been cross-posted to MO last June $10, 2022$.) Let $N = q^k n^2$ be an odd perfect number with special prime $q$...
Jose Arnaldo Bebita Dris's user avatar
1 vote
0 answers
140 views

Why is this alternating summation involving reciprocals of divisors always positive? A conjecture.

Conjecture. For all $n,m \in \Bbb{N}$, $$ f(n, m) := \sum_{c\mid d\mid n\# \\ \gcd(c, 2m) = 1}\dfrac{(-1)^{\omega(d)}}{d} $$ is greater than $0$. Example verification code: ...
SeekingAMathGeekGirlfriend's user avatar
0 votes
4 answers
100 views

Does $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < 10$ hold in general for an odd perfect number $q^k n^2$ with special prime $q$?

This question is an offshoot of this MSE answer. Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. (Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.) If $\sigma(M) = 2M$...
Jose Arnaldo Bebita Dris's user avatar
1 vote
3 answers
115 views

Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$?

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
108 views

Bounds for the abundancy index of divisors of odd perfect numbers in terms of the deficiency function - Part II

This post is an offshoot of this MSE question. Motivation Let $x, y$ and $z$ be positive integers. Denote the sum of divisors of $x$ by $\sigma(x)$. Also, denote the deficiency of $y$ by $D(y)=2y-\...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
135 views

On the quantity $I(q^k) + I(n^2)$ considered as functions of $q$ and $k$, when $q^k n^2$ is an odd perfect number with special prime $q$ - Part II

Hereinafter, let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the abundancy index of the positive integer $x$ ...
Jose Arnaldo Bebita Dris's user avatar
0 votes
3 answers
193 views

On the conjectured inequality $q > k$, where $q^k n^2$ is an odd perfect number with special prime $q$

Let $N=q^k n^2$ be an odd perfect number with special/Eulerian prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
121 views

On the inequality $I(q^k)+I(n^2) \leq \frac{3q^{2k} + 2q^k + 1}{q^k (q^k + 1)}$ where $q^k n^2$ is an odd perfect number

(Note: This post is an offshoot of this earlier MSE question.) Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Define the ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
2 answers
68 views

Is it possible to improve on these bounds for $\frac{\varphi(n)}{n}$, if $q^k n^2$ is an odd perfect number with special prime $q$?

Let $N = q^k n^2$ be an odd perfect number with special/Eulerian prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the Euler-totient function of the positive integer $x$ by ...
Jose Arnaldo Bebita Dris's user avatar
0 votes
0 answers
173 views

A proposed unconditional factor-chain proof for the inequality $I(n) > \frac{3}{2}$, where $q^k n^2$ is an odd perfect number with special prime $q$

Let $N = q^k n^2$ be a hypothetical odd perfect number given in Eulerian form. (That is, $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.) Denote the ...
Jose Arnaldo Bebita Dris's user avatar

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