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Questions tagged [perfect-numbers]

Questions about or involving perfect numbers which are positive integers that are equal to the sum of their proper positive divisors.

0 votes
0 answers
23 views

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 ...
Jose Arnaldo Bebita Dris's user avatar
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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
2 votes
1 answer
69 views

Generalized "perfect numbers" using different n,k values of divisorSum[n, k]

Using the divisor_sigma[n, k] function from the python sympy library where n is the positive integer which is having its divisors added and k is the power each factor is raised to, I was looking for ...
unnamed's user avatar
  • 31
0 votes
1 answer
71 views

prove that $n - \phi(n)$ is a square where $n$ is an even perfect number

where even perfect numbers are of the form $2^{p-1}(2^{p} - 1)$ ( $p$ and $2^{p} - 1$ are prime numbers ) My attempt $\phi(n)$ = ($2^{p - 1} - 2$)($2^{p} - 2$) So, we need to prove that. $2^{p - 1}$($...
Oppenheimer's user avatar
5 votes
0 answers
71 views

Can we efficiently check whether a number is a Zumkeller number?

A positive integer $n$ is a Zumkeller number iff its divisors can be partitioned into two sets with equal sum. If $\sigma(n)$ denotes the divisor-sum-function , this means that there are distinct ...
Peter's user avatar
  • 85.1k
0 votes
1 answer
178 views

Can we really be sure that there is no odd perfect number below $10^{3000}$?

A positive integer $N$ is called perfect if the sum of its divisors (including $1$ and $N$) is $2N$. A famous open problem is whether there is an odd perfect number. Can someone confirm the following ...
Peter's user avatar
  • 85.1k
2 votes
0 answers
62 views

patterns in the abundancy index of integers

Let $\sigma(n)$ be the sum of all divisors (including 1 and $n$) of $n$, and define the abundancy index of $n$ as $I(n) = \sigma(n)/n $. For example: $I(6)= \frac{1+2+3+6}{6} = 1/1+1/2 +1/3 +1/6 = 2$. ...
AndroidBeginner's user avatar
3 votes
1 answer
58 views

Why is this inequality in Brent and Cohens paper on odd perfect numbers true?

In Brent and Cohen's paper about odd perfect numbers, they show this inequality. $N \ge p^a\sigma(p^a) \gt p ^ {2a}$ where a is even. I understand the next second half of this: $p^a\sigma(p^a) \gt p ^ ...
louis's user avatar
  • 45
4 votes
1 answer
152 views

Can an Odd Perfect Number be a Carmichael Number?

Can an odd perfect number $n$ be a Carmichael number? We know that all Carmichael numbers are odd and square-free. But is there a Carmichael number that is also a perfect number? We all know that if ...
Thirdy Yabata's user avatar
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
0 votes
1 answer
106 views

Is any number one less than the sum of its proper divisors?

I've been fascinated by perfect numbers ever since I learned about them, but I don't think 1 should be counted in the sum. Every integer is divisible by itself and 1, so why is the number itself ...
Cain Goldhardt's user avatar
1 vote
1 answer
219 views

Is $28$ the only perfect number that is form of $n^n+1$?

Is $28$ the only perfect number that is form of $n^n+1$? I noticed that in the sequence of $n^n+1$, $28$ is the only perfect number that is in the list. Using Pari GP, I checked the values of $n\leq10^...
Thirdy Yabata's user avatar
0 votes
1 answer
50 views

Specific Prime Numbers Chain proofs of maximum length limits.

Considering Prime Numbers $p\in P$ such that $(p^2+4)\in P, (p^2+4)^2+4\in P, ((p^2+4)^2+4)^2+4\in P, (((p^2+4)^2+4)^2+4)^2+4\in P,((((p^2+4)^2+4)^2+4)^2+4)^2+4\in P$, seems to be either very rare, or ...
Eugen's user avatar
  • 238
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
1 vote
1 answer
394 views

Under what conditions is it true that $m^2 - a$ is not a square if and only if $(m - 1)^2 < m^2 - a < m^2$, where $a>0$?

Preamble: The present inquiry is an offshoot of this MSE question from February 21, 2019, and ultimately, Theorem III.2, page 2 from a paper submitted to a conference organized by De La Salle ...
Jose Arnaldo Bebita Dris's user avatar

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