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Does the following GCD divisibility constraint imply that $\sigma(m^2)/p^k \mid m$, if $p^k m^2$ is an odd perfect number with special prime $p$?

The topic of odd perfect numbers likely needs no introduction. In what follows, denote the classical sum of divisors of the positive integer $x$ by $$\sigma(x)=\sigma_1(x).$$ Let $p^k m^2$ be an odd ...
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0 answers
42 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's user avatar
2 votes
2 answers
193 views

Does an odd perfect number have a divisor (other than $1$) which must necessarily be almost perfect?

Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$. Denote the aliquot sum of $x$ by $s(x)=\sigma(x)-x$ and the deficiency of $x$ by $d(x)=2x-\sigma(x)$. ...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
122 views

Is this disproof for the Descartes-Frenicle-Sorli Conjecture that $k=1$, if $p^k m^2$ is an odd perfect number, valid?

Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. It is known that $$D(p^k)D(m^2)=2s(p^k)s(m^2) \tag{0}$$ where $D(x)=2x-\...
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
153 views

Why is it that, if there are no odd perfect numbers, then there are no other $3$-perfect numbers, apart from the six known, as of the year $1643$?

Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$. A number $N$ is said to be $k$-perfect if $\sigma(N)=kN$ where $k$ is a positive integer. The number $1$ ...
Jose Arnaldo Bebita's user avatar
0 votes
1 answer
70 views

On the prime factorization of $n$ and the quantity $J = \frac{n}{\gcd(n,\sigma(q^k)/2)}$, where $q^k n^2$ is an odd perfect number

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 classical sum of divisors of the positive integer $x$ by $\...
Jose Arnaldo Bebita's user avatar
-1 votes
1 answer
99 views

On a consequence of $G \mid I \iff \gcd(G, I) = G$ (Re: Odd Perfect Numbers and GCDs)

Let $N = q^k n^2$ be an odd perfect number given in the so-called Eulerian form, where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the classical sum of ...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
225 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's user avatar
1 vote
1 answer
134 views

On the equation $s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)}$, if $q^k n^2$ is an odd perfect number with special prime $q$

Let $N$ be an odd perfect number given in the so-called Eulerian form $$N = q^k n^2$$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. In what follows, let ...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
72 views

Does $\sigma(n^2)/q \mid q^k n^2$ imply $\sigma(n^2)/q \mid n^2$, if $q^k n^2$ is an odd perfect number with special prime $q$? - Part II

(Preamble: This inquiry is an offshoot of this MSE question.) Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ as $I(x)=\...
Jose Arnaldo Bebita's user avatar
0 votes
1 answer
77 views

Are these valid proofs for the equation $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$, 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 $q \equiv k \equiv 1 \pmod 4$. It is known that $$i(q)=\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
122 views

Are the (Bezout) coefficients for $\gcd(n^2, \sigma(n^2)) = q\sigma(n^2) - 2(q - 1)n^2$ (where $q^k n^2$ is an odd perfect number) unique?

In what follows, denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the aliquot sum of $x$ by $s(x)=\sigma(x)...
Jose Arnaldo Bebita's user avatar
0 votes
1 answer
88 views

Is the argument used in this proof that $k=1$ logically sound, where $q^k n^2$ is an odd perfect number with special prime $q$?

The topic of odd perfect numbers likely needs no introduction. Euler proved that a hypothetical odd perfect number $N$, if one exists, must have the so-called Eulerian form $N=q^k n^2$, where $q$ is ...
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
121 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's user avatar
1 vote
0 answers
77 views

Follow-up to MSE question 3738458

This is a follow-up inquiry to this MSE question. Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. A number $M$ is said to be perfect if $\sigma(M)=2M$. For example, $6$ and $...
Jose Arnaldo Bebita's user avatar
0 votes
1 answer
101 views

What conditions on $X$ will guarantee that $\gcd(\text{square part of } X,\text{squarefree part of } X)=1$, if $X$ is neither a square nor squarefree?

The following query is an offshoot of this answer to a closely related post. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. The topic of odd perfect ...
Jose Arnaldo Bebita's user avatar
0 votes
0 answers
96 views

Does $I = \gcd(n,\sigma(n^2)) = (\frac{n}{\sigma(q^k)/2})\cdot\gcd(\sigma(q^k)/2,n)$ imply that $\sigma(q^k)/2 \mid n$ holds?

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 GCDs: $$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$ $$H = \...
Jose Arnaldo Bebita's user avatar
0 votes
2 answers
430 views

Does $\sigma(m^2)/p^k$ divide $m^2 - p^k$, if $p^k m^2$ is an odd perfect number with special prime $p$?

The following query is an offshoot of this post 1 and this post 2. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. The topic of odd perfect numbers likely ...
Jose Arnaldo Bebita's user avatar
0 votes
2 answers
327 views

What are the remaining cases to consider for this problem, specifically all the possible premises for $i(q)$?

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. (Note that the divisor sum $\sigma$ is a multiplicative function.) A number $P$ is said to be perfect if $\...
Jose Arnaldo Bebita's user avatar
0 votes
4 answers
102 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's user avatar
1 vote
0 answers
75 views

On the reciprocal of $I(n^2) - \frac{2(q - 1)}{q}$, if $q^k n^2$ is a(n) (odd) perfect number with special prime $q$

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. A number $N$ is said to be perfect if $\sigma(N)=...
Jose Arnaldo Bebita'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's user avatar
1 vote
1 answer
110 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's user avatar
3 votes
1 answer
130 views

If $\gcd(n^2, \sigma(n^2)) = q\sigma(n^2) - 2(q - 1) n^2$, does it follow that $q n^2$ is perfect?

Let $q$ be an (odd) prime, and let $\gcd(q,n)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. A number $N$ is said to be perfect if $\sigma(N)=2N$. ...
Jose Arnaldo Bebita'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's user avatar
3 votes
2 answers
158 views

Help with "A Simpler Dense Proof regarding the Abundancy Index."

I'm reading Richard Ryan's article "A Simpler Dense Proof regarding the Abundancy Index" and got stuck in his proof for Theorem 2. The Theorem is stated as follows: Suppose we have a ...
jvkloc's user avatar
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2 votes
0 answers
107 views

If $k=1$, then the index of an odd perfect number $p^k n^2$ (as defined by Chen and Chen (2014)) is not squarefree.

In what follows, we let $\sigma=\sigma_1$ denote the classical sum-of-divisors function. Denote the deficiency function of the positive integer $x$ by $D(x)=2(x)-\sigma(x)$, and the aliquot sum of $x$...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
123 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's user avatar
2 votes
1 answer
178 views

On odd perfect numbers $q^k n^2$ and the deficient-perfect divisor $q^{\frac{k-1}{2}} n^2$ - Part III

(Preamble: This question is an offshoot of this answer by mathlove to an earlier post.) Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of the divisors of the positive integer $x$. If $\sigma(m)=...
Jose Arnaldo Bebita's user avatar
2 votes
2 answers
97 views

On improving the upper bound $I(m^2) \leq \frac{2p}{p+1}$, if $p^k m^2$ is an odd perfect number with special prime $p$

(Preamble: This question is an offshoot of this earlier MSE post.) Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. ...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
110 views

Will it be possible to compute a factored expression for $n^2 - q^k$, if $q^k n^2$ is an odd perfect number with special prime $q$?

In what follows, we denote the classical sum of divisors of the positive integer $x$ by $$\sigma(x)=\sigma_1(x)=\sum_{d \mid x}{d},$$ and the abundancy index of $x$ by $I(x)=\sigma(x)/x$. If $N$ is ...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
119 views

On the inequality $I(n^2) \geq (6 - s(q^k))/3$, where $q^k n^2$ is an odd perfect number and $I(x)$ is the abundancy index of $x$

In what follows, we denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, the aliquot sum of $x$ by $s(x)=\sigma(x)-...
Jose Arnaldo Bebita's user avatar
0 votes
1 answer
71 views

On an inequality involving the abundancy index of the Eulerian component $p^k$ of an odd perfect number $p^k m^2$ - Part II

Preamble: This post is an offshoot of this earlier question. The topic of odd perfect numbers likely needs no introduction. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
124 views

If $N = q^k n^2$ is an odd perfect number with special prime $q$, then must $\sigma(n^2)$ be abundant?

Preamble: This post is an offshoot of this earlier MSE question. The topic of odd perfect numbers likely needs no introduction. Let $\sigma=\sigma_{1}$ denote the classical sum of divisors. Denote ...
Jose Arnaldo Bebita's user avatar
5 votes
1 answer
506 views

Proof-verification request: If $q^k n^2$ is an odd perfect number with special prime $q$, then $q < n^{2/5}$.

Hereinafter, we will denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. The topic of odd perfect numbers likely needs no introduction. Let $q^k n^2$ be an odd ...
Jose Arnaldo Bebita's user avatar
2 votes
2 answers
182 views

Is this proof regarding the nonexistence of odd perfect numbers correct?

Preamble: This post is an offshoot of this earlier question, which was not so well-received in MathOverflow. Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
134 views

Is it true that $\frac{D(n^2)}{s(q^k)} > \frac{\sigma(q^k)}{2}$, if $q^k n^2$ is an odd perfect number given in Eulerian form?

Preamble: This post is an attempt to resolve the problem in this earlier question. Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. A number $M$ is said to be perfect if $\...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
73 views

On the equation $\frac{1}{q}=1-\frac{\varphi(N)/N}{\varphi(n)/n}$, where $N=q^k n^2$ is an odd perfect number with special prime $q$

Preamble: This question is an offshoot of this earlier MSE post. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, and the Euler totient function of $x$ by $\...
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
76 views

On the sum $I(q^k) + I(n^2)$, where $q^k n^2$ is an odd perfect number with special prime $q$

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, and denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. Let $N = q^k n^2$ be an odd perfect number ...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
127 views

Does $\sigma(n^2)/q \mid q^k n^2$ imply $\sigma(n^2)/q \mid n^2$, if $q^k n^2$ is an odd perfect number with special prime $q$?

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$. I discovered an interesting identity involving ...
Jose Arnaldo Bebita's user avatar
0 votes
3 answers
525 views

Does the following lower bound improve on $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number? - Part II

Preamble: This question is an offshoot of these earlier posts: (post1), (post2). 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,...
Jose Arnaldo Bebita's user avatar
2 votes
2 answers
220 views

If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m)$?

(Preamble: This question is tangentially related to this earlier post.) Denote the classical sum of divisors of the positive integer $x$ to be $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $...
Jose Arnaldo Bebita's user avatar
2 votes
0 answers
50 views

Improving $I(m) < 2$, if $p^k m^2$ is an odd perfect number with special prime $p$

The topic of odd perfect numbers likely needs no introduction. Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$, and let $I(x)=\sigma(x)/x$ denote the ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
119 views

On an inequality involving the abundancy index of the Eulerian component $p^k$ of an odd perfect number $p^k m^2$

The topic of odd perfect numbers likely needs no introduction. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the abundancy index of $x$ by $I(x)=\sigma(x)...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
137 views

If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m^2)$?

Denote the classical sum of divisors of the positive integer $x$ to be $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. Finally, denote the deficiency of $x$ by $D(x)...
Jose Arnaldo Bebita's user avatar
3 votes
0 answers
232 views

On odd perfect numbers $p^k m^2$ with special prime $p$, satisfying $m^2 - p^k = 2^r t$

In what follows, we will denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. A positive integer $y$ ...
Jose Arnaldo Bebita's user avatar
2 votes
2 answers
123 views

If $p^k m^2$ is an odd perfect number, then is there a constant $D$ such that $\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}$?

(Note: This question is an offshoot of this closely related one.) Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. The topic of odd perfect numbers likely ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
66 views

Follow-up to question 3121498, asked in February 2019

Let $n = p^k m^2$ be an odd perfect number given in Eulerian form (i.e. $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$). Denote the classical sum of ...
Jose Arnaldo Bebita's user avatar
1 vote
1 answer
70 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 $q$

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's user avatar
2 votes
3 answers
128 views

Applying a criterion on deficient numbers to the proper factors of an odd perfect number

Hereinafter, 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 $x$. Denote the deficiency of $x$ by $$D(...
Jose Arnaldo Bebita's user avatar