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0 votes
0 answers
72 views

Is there any square harmonic divisor number greater than $1$?

A harmonic divisor number or Ore number is a positive integer whose harmonic mean of its divisors is an integer. In other words, $n$ is a harmonic divisor number if and only if $\dfrac{nd(n)}{\sigma(n)...
Jianing Song's user avatar
  • 1,923
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
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 Dris'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 Dris's user avatar
0 votes
2 answers
324 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 Dris's user avatar
1 vote
1 answer
52 views

Question about a result on odd perfect numbers - Part II

(This question is an offshoot of this earlier one.) In the paper titled Improving the Chen and Chen result for odd perfect numbers (Lemma 8, page 7), Broughan et al. show that if $$\frac{\sigma(n^2)}{...
Jose Arnaldo Bebita Dris's user avatar
7 votes
2 answers
117 views

If $q$ is prime, can $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ be both squares when $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$?

This is related to this earlier MSE question. In particular, it appears that there is already a proof for the equivalence $$\sigma(q^{k-1}) \text{ is a square } \iff k = 1.$$ Let $\sigma(x)$ ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
2 answers
69 views

If $q$ is a Fermat prime, is $\sigma(q^k)/2$ a square if $k \equiv 1 \pmod 4$?

In what follows, let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. A number of the form $M = 2^{2^m} + 1$ is called a Fermat number. If in addition $M$ is prime, then $M$ is ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
0 answers
77 views

A consequence of assuming the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers

Let $x$ be a positive integer. Denote the sum of divisors of $x$ by $$\sigma(x) = \sum_{d \mid x}{d},$$ and the deficiency of $x$ by $$D(x) = 2x - \sigma(x).$$ A number $N$ is said to be perfect if $...
Jose Arnaldo Bebita Dris's user avatar
2 votes
1 answer
271 views

Odd abundant numbers and the condition $\sqrt{n^2+12n\sigma(n)}\in\mathbb{Z}_{\geq 1}$

We denote for integers $n\geq 1$ the sum of divisors function as $\sigma(n)=\sum_{d\mid n}d$. If we assume that $m$ is an odd perfect number then it satisfies that $m^2+12m\sigma(m)$ is a perfect ...
user avatar
10 votes
2 answers
872 views

A conjecture concerning the number of divisors and the sum of divisors.

I stumbled upon the following conjecture: Let $n$ be a positive integer. Let $\sigma\left(n\right)$ be the sum of all (positive) divisors of $n$, and let $\tau\left(n\right)$ be the number of these ...
user avatar
2 votes
1 answer
744 views

Sum of divisors is power of 2, is n square-free?

We were given this question on one of our recent exams, and we can't seem to generate a proof, nor find a counter-example. If we let $\sigma(n)$ be the sum of the divisors of $n$, the question is, if ...
superckl's user avatar
  • 797
1 vote
1 answer
47 views

If $\sigma _{1}(n)\mid \sigma _{2}(n)$, does $n$ has to be a perfect square?

Let's say $\sigma _{1}(n)\mid \sigma _{2}(n)$. Can we say, therefore $n$ has to be a perfect square? How to show that?
esege's user avatar
  • 3,641
1 vote
1 answer
46 views

Are there infinitely many $k$ for which $\frac{\sigma(k)}{k}$ is a rational square where $ \sigma(k) $ and $k$ both are square?

Is There some one who can show me if there are infinitely many $k$ for which $$\frac{\sigma(k)}{k}$$ is a rational square where $\sigma(k)$ and $k$ both are square ? Note :$\sigma(k)$ is sum ...
zeraoulia rafik's user avatar
10 votes
2 answers
2k views

When is the sum of divisors a perfect square?

For $n=3$, $\sigma(n)=4$, a perfect square. Calculating further was not yielding positive results. I was wondering is there a way to find all such an $n$, like some algorithm? We know that if $n=p_1^...
Bhaskar Vashishth's user avatar

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