In what follows, we let $\sigma(X)$ denote the sum of the divisors of the positive integer $X$. Denote the abundancy index of $X$ by $I(X)=\sigma(X)/X$, and the deficiency of $X$ by $D(X)=2X-\sigma(X)$. Finally, let $s(X)=\sigma(X)-X$ denote the sum of the aliquot divisors of $X$.

This is a question about Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College, titled Characterization of the Positive Integers with Abundancy Index of the Form $(2x-1)/x$. (A copy of the poster presentation is available via the following hyperlink.)

In the abstract of the paper, it is stated in the fourth sentence that

Rational numbers of the form $(2x-1)/x$ are important since both even and odd perfect numbers have a divisor with abundancy index of this form.

Let $M = 2^{p-1}(2^p - 1)$ be an even perfect number, and let $N = q^k n^2$ be an odd perfect number.

Clearly, $$I(2^{p-1}) = \frac{2^p - 1}{2^{p-1}} = \frac{2x_1 - 1}{x_1}$$ where $x_1 = 2^{p-1}$. (In other words, $2^{p-1}$ is an even almost perfect number, since it is a power of two.)

However, $$I(p^k) = \frac{p^{k+1} - 1}{p^{k+1} - p^k}$$ and $$I(n^2) = \frac{2}{I(p^k)} = \frac{2(p^{k+1} - p^k)}{p^{k+1} - 1}$$ so clearly $p^k$ is not almost perfect (since $p$ must be odd).

Additionally, since $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)} \geq 3$$ (see the paper [Dris, 2012]), then clearly $n^2$ is likewise not almost perfect. (Similarly, it can be proved that $n$ and $q^k n$ are not almost perfect.)

So I think the trivial divisor $1$ of an odd perfect number has the required abundancy index $$I(1) = 1 = \frac{2\cdot{1} - 1}{1} = \frac{2x_2 - 1}{x_2}$$ where $x_2 = 1$.

Here is my question:

Is there any other divisor $m > 1$ of an odd perfect number $N = q^k n^2$ such that $$I(m) = \frac{2x - 1}{x}$$ for some positive integer $x$?

  • $\begingroup$ This looks like it probably should be on StackOverFlow $\endgroup$
    – IAmNoOne
    Jan 3 '19 at 13:04
  • 1
    $\begingroup$ @IAmNoOne, did you mean MathOverflow? $\endgroup$ Jan 3 '19 at 19:50

Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$.

Suppose that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.

Then $$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} = 2 - \frac{1}{(q+1)/2} = \frac{2((q+1)/2) - 1}{(q+1)/2},$$ where $(q+1)/2$ is an integer (since $q \equiv 1 \pmod 4$).

It remains to consider the case when $k>1$. (Note that $k \equiv 1 \pmod 4$.)


The following answer is based on a theorem in a paper (to appear) communicated to Dris by Holdener (co-authored with Rachfal).

Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$. Suppose that $k>1$.

Consider the proper factor $$q^{\frac{k-1}{2}} n^2.$$

This has abundancy index $$I\bigg(q^{\frac{k-1}{2}} n^2\bigg) = I\bigg(q^{\frac{k-1}{2}}\bigg)I(n^2) = I\bigg(q^{\frac{k-1}{2}}\bigg)\cdot\frac{2}{I(q^k)} = \frac{q^{\frac{k+1}{2}} - 1}{q^{\frac{k-1}{2}}(q - 1)}\cdot\frac{2q^k (q - 1)}{q^{k+1} - 1}.$$

Since $k \equiv 1 \pmod 4$, $k+1$ is even, so that $$q^{k+1} - 1 = \bigg(q^{\frac{k+1}{2}} - 1\bigg)\cdot\bigg(q^{\frac{k+1}{2}} + 1\bigg).$$

Canceling the common factor $$q^{\frac{k-1}{2}} (q - 1)\bigg(q^{\frac{k+1}{2}} - 1\bigg)$$ in the numerator and denominator of $$I\bigg(q^{\frac{k-1}{2}} n^2\bigg),$$ we get $$I\bigg(q^{\frac{k-1}{2}} n^2\bigg) = \frac{2q^{\frac{k+1}{2}}}{q^{\frac{k+1}{2}} + 1} = 2 - \frac{1}{\bigg(\frac{q^{\frac{k+1}{2}}+1}{2}\bigg)},$$ as desired.


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