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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 ...
Jose Arnaldo Bebita's user avatar
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 ...
Jose Arnaldo Bebita's user avatar
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 ...
Jose Arnaldo Bebita's user avatar
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 ...
Jose Arnaldo Bebita's user avatar