Skip to main content

All Questions

1 vote
1 answer
323 views

Who discovered the largest known $3$-perfect number in $1643$?

Multiperfect numbers probably need no introduction. (These numbers are defined in Wikipedia and MathWorld.) I need the answer to the following question as additional context for a research article ...
Jose Arnaldo Bebita Dris's user avatar
2 votes
2 answers
183 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 Dris's user avatar
4 votes
1 answer
278 views

Determining whether $\sigma(q^k)/2$ is squarefree, where $q^k n^2$ is an odd perfect number with special prime $q$

Preamble: The present inquiry is an offshoot of What are the remaining cases to consider for this problem, specifically all the possible premises for $i(q)$?. MOTIVATION Denote the classical sum of ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
137 views

If $p^k m^2$ is an odd perfect number with special prime $p$ and $p = k$, then $\sigma(p^k)/2$ is not squarefree.

While researching the topic of odd perfect numbers, we came across the following implication, which we currently do not know how to prove: CONJECTURE: If $p^k m^2$ is an odd perfect number with ...
Jose Arnaldo Bebita Dris's user avatar
4 votes
0 answers
150 views

Is $650$ the only solution not fitting in the family?

Inspired by this question The linked question conjectures that $\frac{\sigma(n)}{n+1}$ (where $\sigma(n)$ denotes the divisor-sum function) is not an integer for any squarefree composite number. If we ...
Peter's user avatar
  • 85.1k
1 vote
1 answer
132 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 Dris'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 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
2 votes
2 answers
173 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 Dris's user avatar
2 votes
1 answer
130 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 Dris's user avatar
1 vote
1 answer
118 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 Dris's user avatar
0 votes
3 answers
521 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 Dris'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 Dris's user avatar
0 votes
1 answer
67 views

A proof (?) for $k = 1 \implies q \neq 5$, 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 given in Eulerian form (i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$). Inspired by mathlove's answer to ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
71 views

Is there an odd $x$ such that $2x^2 \equiv 0 \pmod {\sigma(x)}$ and $\sigma(x^2) \equiv 0 \pmod {\sigma(x) - 1}$?

CONTEXT This question is a result of considerations stemming from this closely related MO question. INITIAL QUESTION My question is as is in the title: Is there an odd $x$ such that $$2x^2 \equiv 0 \...
Jose Arnaldo Bebita Dris's user avatar

15 30 50 per page
1
2 3 4 5
7