All Questions
6 questions
6
votes
1
answer
2k
views
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 ...
CommunityBot
- 1
2
votes
2
answers
484
views
On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II
(Preamble: We have asked this same question in MSE two weeks ago, without getting any answers. We have therefore cross-posted it to MO, hoping that it gets answered here.)
The topic of odd perfect ...
0
votes
0
answers
107
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 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 ...
2
votes
1
answer
259
views
On attempting a proof for $r > 1$, if $M = {2^r}{b^2}$ is an even almost perfect number which is not a power of two
(Preamble: I first thought that this question might be more appropriate for MSE. However, I posted it here nonetheless in the hope that someone with that brilliant idea can help with answering my ...
CommunityBot
- 1
1
vote
0
answers
256
views
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 ...
CommunityBot
- 1
2
votes
0
answers
286
views
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 ...