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(Note: This has been cross-posted to MSE. However, I feel that it is more likely to receive a good answer here, because I believe that it is a research-level question. For the mathematicians who voted to put this question on hold, I invite you to peruse this preprint to give you an inkling of where I'm driving at. Thanks!)

Let $\sigma$ be the classical sum-of-divisors function.

A number is said to be perfect if $\sigma(N)=2N$.

If $q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$), are the following statements known to hold in general?

$\bf{(a)}$ $\gcd(n^2, \sigma(n^2))$ is large.

$\bf{(b)}$ The deficiency $D(n^2) = 2n^2 - \sigma(n^2)$ is large.

$\bf{(c)}$ The index $i(q^k) = \sigma(N/q^k)/q^k$ is large.

Using the trivial relationships:

$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = i(q^k) = \frac{2n^2}{\sigma(q^k)}$$

then if the Descartes-Frenicle-Sorli conjecture that $k = 1$ is true, it is possible to show that a lower bound for the quantities $\bf{(a)}$, $\bf{(b)}$ and $\bf{(c)}$ is given by $n/\sqrt{3}$. (Here, I have used Acquaah and Konyagin's estimate $q < n\sqrt{3}$. The inequality $q^k < n^2$ then gives the desired large numerical bound if we use known lower bounds for the odd perfect number $N = q^k n^2$, latest of which are by Ochem and Rao.)

What happens when $k > 1$? I do know that $\sigma(n^2)/q^k \geq 315$ by using a result of Broughan, Delbourgo, and Zhou.

Is it possible to do better than this, apart from attempting a proof of (obviously) $q^k < n$?

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closed as off-topic by Daniel Loughran, Stefan Kohl, Dima Pasechnik, Lucia, Alex Degtyarev May 26 '15 at 6:35

This question appears to be off-topic. The users who voted to close gave these specific reasons:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Stefan Kohl, Dima Pasechnik, Alex Degtyarev
  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Daniel Loughran, Lucia
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Why are people voting to close this question? =( $\endgroup$ – Jose Arnaldo Bebita-Dris May 26 '15 at 0:28
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If $k > 1$, we can do better if the following inequalities hold:

$$q < q^k < n.$$

The resulting lower bound is

$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = i(q^k) = \frac{2n^2}{\sigma(q^k)} > \frac{8}{5}\cdot\frac{n^2}{q^k} > \frac{8}{5}\cdot{n}.$$

Since $k > 1$ implies $q < n$, it remains to consider the case

$$q < n < q^k.$$

(Update (February 2016): In a recent preprint, Brown claims a complete proof for $q < n$, and a partial proof for the inequality $q^k < n$. If completed, Brown's proof then rules out this last remaining case.)

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