# Can this bound for the abundancy index of $n$ be improved, given that $q^k n^2$ is an odd perfect number with $k=1$?

In what follows, set $$I(x)=\sigma(x)/x$$ to be the abundancy index of $$x \in \mathbb{N}$$, where $$\sigma(x)$$ is the sum of divisors of $$x$$. If $$I(y)=2$$ and $$y$$ is odd, then $$y$$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $$N = q^k n^2$$ where $$q$$ is the special/Euler prime satisfying $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$.

The following result was communicated to me (via e-mail, by Pascal Ochem) on April 17, 2013.

If $$N = {q^k}{n^2}$$ is an odd perfect number given in Eulerian form, then $$I(n) > {\left(\frac{8}{5}\right)}^{\frac{\ln(4/3)}{\ln(13/9)}} \approx 1.44440557.$$

The proof of this result uses the following lemma.

Let $$x(n) = \ln(I(n^2))/\ln(I(n))$$. If $$\gcd(a, b) = 1$$, then $$\min(x(a), x(b)) < x(ab) < \max(x(a),x(b)).$$

Proof of Lemma: First, note that $$x(a) \neq x(b)$$ (since $$\gcd(a , b) = 1$$). Without loss of generality, we may assume that $$x(a) < x(b)$$. Thus, since $$x(n) = \frac{\ln(I(n^2))}{\ln(I(n))},$$ we have $$\frac{\ln(I(a^2))}{\ln(I(a))} < \frac{\ln(I(b^2))}{\ln(I(b))}.$$ This implies that $$\ln(I(a^2))\ln(I(b)) < \ln(I(b^2))\ln(I(a)).$$ Adding $$\ln(I(a^2))\ln(I(a))$$ to both sides of the last inequality, we get $$\ln(I(a^2))\bigg(\ln(I(b)) + \ln(I(a))\bigg) < \ln(I(a))\bigg(\ln(I(b^2)) + \ln(I(a^2))\bigg).$$ Using the identity $$\ln(X) + \ln(Y) = \ln(XY)$$, we can rewrite the last inequality as $$\ln(I(a^2))\ln(I(ab)) < \ln(I(a))\ln(I((ab)^2))$$ since $$I(x)$$ is a weakly multiplicative function of $$x$$ and $$\gcd(a, b) = 1$$. It follows that $$\min(x(a), x(b)) = x(a) = \frac{\ln(I(a^2))}{\ln(I(a))} < \frac{\ln(I((ab)^2))}{\ln(I(ab))} = x(ab).$$

Under the same assumption $$x(a) < x(b)$$, we can show that $$x(ab) < x(b) = \max(x(a),x(b))$$ by adding $$\ln(I(b^2))\ln(I(b))$$ ( instead of $$\ln(I(a^2))\ln(I(a))$$ ) to both sides of the inequality $$\ln(I(a^2))\ln(I(b)) < \ln(I(b^2))\ln(I(a)).$$

This finishes the proof.

From this lemma, we note that $$1 < x(n) < 2$$ follows from $$I(n) < I(n^2) < {\left(I(n)\right)}^2$$ and $$I(n^2) = {\left(I(n)\right)}^{x(n)}$$.

The trivial lower bound for $$I(n)$$ is $${\bigg(\frac{8}{5}\bigg)}^{1/2} < I(n).$$

Note that decreasing the denominator in the exponent gives an increase in the lower bound for $$I(n)$$.

Proof of Result: We sketch a proof for the mentioned result here, as communicated to the author by Pascal Ochem.

Suppose that $$N = {q^k}{n^2}$$ is an odd perfect number given in Eulerian form.

We want to obtain a lower bound on $$I(n)$$. We know that $$I(n^2) = 2/I(q^k) > 2/(5/4) = 8/5.$$ We need to improve the trivial bound $$I(n^2) < {\left(I(n)\right)}^2$$.

Let $$x(n)$$ be such that $$I(n^2) = {\bigg(I(n)\bigg)}^{x(n)}.$$

That is, $$x(n) = \ln(I(n^2))/\ln(I(n))$$. We want an upper bound on $$x(n)$$ for $$n$$ odd. By the lemma, we consider the component $$r^s$$ with $$r$$ prime that maximizes $$x(r^s)$$.

We have $$I(r^s) = \frac{r^{s + 1} - 1}{{r^s}(r - 1)} = 1 + \frac{1}{r - 1} - \frac{1}{{r^s}(r - 1)}.$$ Also, $$I(r^{2s}) = \frac{r^{2s + 1} - 1}{{r^{2s}}(r - 1)} = I(r^s)\bigg(1 + \left(\frac{1 - r^{-s}}{r^{s + 1} - 1}\right)\bigg).$$ So, $$x(r^s) = \frac{\ln(I(r^{2s}))}{\ln(I(r^s))} = \frac{\ln(I(r^s)) + \ln(1 + \left(\frac{1 - r^{-s}}{r^{s + 1} - 1}\right))}{\ln(I(r^s))},$$ from which it follows that $$x(r^s) = 1 + \frac{\ln(1 + \left(\frac{1 - r^{-s}}{r^{s + 1} - 1}\right))}{\ln(1 + \frac{1}{r - 1} - \frac{1}{{r^s}(r - 1)})}.$$ We can check that $$x(r^s) > x(r^t)$$ if $$s < t$$ and $$r \geq 3$$. Therefore, $$x(r^s)$$ is maximized for $$s = 1$$. Now, $$x(r) = 1 + \frac{\ln(1 + (1/(r(r + 1))))}{\ln(1 + (1/r))} = \frac{\ln(1 + (1/r) + (1/r)^2)}{\ln(1 + (1/r))} = \ln(I(r^2))/\ln(I(r)),$$ which is maximized for $$r = 3$$. So, $$x(3) = \ln(I(3^2))/\ln(I(3)) = \ln(13/9)/\ln(4/3) \approx 1.27823.$$

The claimed result then follows, and the proof is complete.

Motivation for My Inquiry

Note that, since $$q$$ is prime with $$q \equiv 1 \pmod 4$$, then $$q \geq 5$$. In particular, when $$k=1$$, then we obtain $$I(q^k) = 1 + \frac{1}{q} \leq \frac{6}{5} \iff I(n^2) = \frac{2}{I(q^k)} \geq \frac{5}{3},$$ from which we get $$I(n) > \bigg(I(n^2)\bigg)^{1/{x(3)}} \geq \bigg(\frac{5}{3}\bigg)^{\frac{\ln(4/3)}{\ln(13/9)}} \approx 1.4912789897463723558.$$

Here is my question:

Would it be possible to tweak the argument in this post in order to come up with a proof for the implication $$k=1 \implies I(n) > \frac{3}{2}?$$

Let $$q^k n^2$$ be an odd perfect number with special/Euler prime $$q$$ satisfying $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$.

First off, we show the following lemma.

Lemma 1 $$I(n) \neq 3/2$$

Proof The proof is trivial and follows from the fact that $$n \neq 2$$, since $$n$$ must be odd. (In fact, we know by Nielsen's latest result that $$\omega(n) \geq 9$$, where $$\omega(n)$$ is the number of distinct prime factors of $$n$$. In particular, this means that $$n$$ is composite.) Note that $$2$$ is solitary as it is prime.

Next, we prove the following implication.

Theorem $$\bigg((k=1) \land (I(n) < 3/2)\bigg) \implies q=5$$

Proof Since $$k=1$$, then we obtain $$\frac{2q}{q+1}=I(n^2) \leq (I(n))^{\frac{\ln(13/9)}{\ln(4/3)}} < \bigg(\frac{3}{2}\bigg)^{\frac{\ln(13/9)}{\ln(4/3)}}.$$ Using WolframAlpha, we get that $$-1 < q < 5.23316$$, from which we conclude that $$q=5$$, since $$q$$ is a prime satisfying $$q \equiv 1 \pmod 4$$.

In general, we can remove the assumption that $$k=1$$:

Theorem $$I(n) < 3/2 \implies q=5$$

Proof We proceed as follows: $$\frac{2(q-1)}{q} < I(n^2) \leq (I(n))^{\frac{\ln(13/9)}{\ln(4/3)}} < \bigg(\frac{3}{2}\bigg)^{\frac{\ln(13/9)}{\ln(4/3)}}.$$

Using WolframAlpha, we obtain $$0 < q < 6.23316$$, which implies that $$q=5$$, since $$q$$ is a prime number satisfying $$q \equiv 1 \pmod 4$$.

Remarks: Note that $$I(n) > 3/2$$ would also follow from the inequality $$D(n) < s(n)$$, where $$D(n)=2n-\sigma(n)$$ is the deficiency of $$n$$ and $$s(n)=\sigma(n) - n$$ is the sum of the aliquot divisors of $$n$$.

• Note that, if $q > 5$, then since $q$ is prime and $q \equiv 1 \pmod 4$, we have that $q \geq 13$, so that we obtain $$\frac{24}{13} \leq \frac{2(q-1)}{q} < I(n^2) < (I(n))^{\frac{\ln(13/9)}{\ln(4/3)}}$$ from which we get $$I(n) > \bigg(\frac{24}{13}\bigg)^{\frac{\ln(4/3)}{\ln(13/9)}} \approx 1.6155087635466.$$ – Jose Arnaldo Bebita-Dris Oct 30 '18 at 12:47
• Notice that we also have the biconditional $$\bigg(D(n) < \frac{n}{2} < s(n)\bigg) \iff I(n) > \frac{3}{2}.$$ – Jose Arnaldo Bebita-Dris Nov 5 '18 at 2:56