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Tagged with divisors-multiples perfect-numbers
22 questions with no upvoted or accepted answers
9
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Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers
In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following recurrence relation ($n>1$):
$$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\...
- 213
6
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1
answer
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If $N = qn^2$ is an odd perfect number with $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$?
The title says it all.
Question
If $N = qn^2$ is an odd perfect number with Euler prime $q$ and $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$?
Heuristic
From the Descartes spoof, with ...
5
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0
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171
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Touchard / van der Pol's identity for the sum of divisors and an elliptic curve for perfect numbers
In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$, satisifies the following recurrence relation ($n>1$):
$$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\...
- 213
4
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87
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On Carmichael function and aliquot parts of odd perfect numbers
I've asked nine months ago this question on Mathematics Stack Exchange with identifier 4430381 and same title. There is not answer for this question on Mathematics Stack Exchange, I wondered if this ...
3
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0
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180
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Does this Theorem 2 from Dandapat et al. imply that $\gcd(\sigma(p^k),\sigma(a^2)) > 1$?
Write the odd perfect number $m=p^k a^2$ as a product of primes
$$m = p^k {p_1}^{2a_1} \cdots {p_v}^{2a_v}.$$
(Note that it is known that $v \geq 9$ by work of Nielsen.) Let $N(m)$ be the number of ...
2
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0
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751
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Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number?
Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$.
It is known that
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))...
2
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0
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342
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On odd perfect numbers and a GCD - Part II
(Note: A detailed version of this question was posted in MSE last April 15, 2020. It has not received any responses there as of yet. I have therefore cross-posted it here, hoping that it is ...
2
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0
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117
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If $n$ is a multiperfect number, then necessarily does one of its prime factors $p$ satisfy $p \parallel n$?
My question is as in the title:
If $n$ is a multiperfect number, then necessarily does one of its prime factors $p$ satisfy $p \parallel n$?
I quote from an answer by Varun Vejalla to a closely ...
2
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0
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76
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Least number of factors $\sigma(p^e)$ of representation of $\sigma(N)$ to get the least multiple of $\operatorname{rad}(N)$, for odd perfect numbers
I've cross-posted this from the post of Mathematics Stack Exchange that I've asked (Apr, 2nd 2020) with title On the least number of factors $\sigma(q^{e_q})$ to get the least multiple of $\...
2
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0
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68
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Is it possible to deduce statements for odd perfect numbers from the convolution sums involving divisor functions or other arithmetic functions?
Dividing and using some identities of [1] I've deduced the following facts, see also the remarks below. After these introductory paragraphs, to motivate our question, I am asking if we can deduce some ...
2
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0
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57
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On $\sum_{\substack{1\leq d\mid n\\d<f(n)}}d$ and odd perfect numbers, for $f(n)$ the greatest prime factor or $\operatorname{rad}(n)$, respectively
First, in this paragraph we remember the definitions/notations for two number theoretic functions, for an integer $m>1$, we denote its greatest prime factor as $\operatorname{gpf}(m)$, and the ...
2
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0
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221
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Determine whether if $n$ is a primitive pseudoperfect (semiperfect) number, then $\sigma(n)<2^{\sigma_0(n)}$
Determine whether if $n$ is a primitive pseudoperfect (semiperfect) number, then $\sigma(n)<2^{\sigma_0(n)}$.
$\sigma_k(n)$ is the division function and $\sigma(n)=\sigma_1(n)$. A number is ...
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2
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What are the divisors of $2n^2 - \sigma_{1}(n^2)$ for composite $n$?
What are the divisors of $2n^2 - \sigma_{1}(n^2)$ for composite $n$?
Here, $\sigma_{1}$ is the classical sum-of-divisors function. For example, $\sigma_{1}(3^2) = 1 + 3 + {3^2} = 13$.
(The function ...
2
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1
answer
482
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If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above?
STATEMENT OF THE PROBLEM
If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above?
MOTIVATION
Let $\sigma=\sigma_{1}$ denote the classical ...
1
vote
0
answers
61
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Is $N - \varphi(N)$ a square, if $N = q^k m^2$ is an odd perfect number with special prime $q$?
This question was inspired by this MSE question.
In MSE, it is shown that
$$n - \varphi(n) = (2^{p-1})^2$$
if $n = {2^{p-1}}(2^p - 1)$ is an even perfect number.
Here is my question in this post:
Is $...
1
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0
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167
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On "Euclidean" odd perfect numbers
In what follows, we let $N = r^s u^2$ be an odd perfect number given in Eulerian form, i.e. $r$ is the special prime satisfying $r \equiv s \equiv 1 \pmod 4$ and $\gcd(r,u)=1$. In this preprint, ...
1
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Equations involving quasiperfect numbers: a first search of odd solutions for this type of equations or well succinct reasonings about these
In this post we study the following equations that involve quasiperfect numbers, denoted as $x$, that are integers such that the sum of all its positive divisors is equals to $2x+1$, and certain ...
1
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Attempt of exploit the equation $1/\operatorname{rad}(n)=1/2-2\varphi(n)/\sigma(n)$ in the context of even perfect numbers, and a related conjecture
It is well known that the problem concerning even perfect numbers is to prove or refute if there are infinitely many of them. Few weeks ago I wrote the following conjecture, where $\varphi(n)$ denotes ...
1
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0
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141
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Is there an integer $r \neq q$ (with $r>1$) such that $N = q^k n^2 = \frac{r(r+1)}{2}\cdot{d}$ is an odd perfect number with $d>1$?
Slowak showed in 1999 that every odd perfect number $N = q^k n^2$ can be written in the form
$$N = \dfrac{{q^k}\sigma(q^k)}{2}\cdot{D}$$
where $D>1$.
From this result, it follows that every odd ...
1
vote
0
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256
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On even almost perfect numbers other than powers of two
(Note: This question is an improved version of and has been cross-posted from this MSE post.)
Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an ...
0
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0
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55
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If $p^k m^2$ is an odd perfect number with special prime $p$, is it possible to have $p = k$?
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
My question is as is in the title:
If $p^k m^2$ is an odd perfect number with special prime $p$, is it ...
0
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0
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107
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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 answer to an MSE question with the same title.)
Denote the classical sum of the divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$ and the ...