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 $p$

Question

(Preamble: This question is an offshoot of this earlier MSE post.)

Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

Note that it is trivial to prove that $$\frac{p+1}{p} \leq I(p^k) < \frac{p}{p-1}$$ from which we obtain $$\frac{2(p-1)}{p} < I(m^2) = \frac{2}{I(p^k)} \leq \frac{2p}{p+1}.$$ This implies that $$\frac{2}{p+1} \leq \frac{D(m^2)}{m^2} < \frac{2}{p}.$$ Taking reciprocals, multiplying by $2$, and subtracting $1$, we get $$p-1 < \frac{\sigma(m^2)}{D(m^2)} \leq p,$$ where we note that $(2/p) < p-1$.

Now consider the quantity $$\bigg(\dfrac{D(m^2)}{m^2} - \dfrac{2}{p}\bigg)\bigg(\dfrac{\sigma(m^2)}{D(m^2)} - \dfrac{2}{p}\bigg).$$ This quantity is negative. Thus, we obtain $$I(m^2) + \bigg(\dfrac{2}{p}\bigg)^2 < \dfrac{2}{p}\bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\bigg) = \dfrac{2}{p}\Bigg(\bigg(2-I(m^2)\bigg) + \bigg(\dfrac{I(m^2)}{2-I(m^2)}\bigg)\Bigg).$$ Now, let $z_1=I(m^2)$. Then we have the inequality $$z_1 + \bigg(\dfrac{2}{p}\bigg)^2 < \dfrac{2}{p}\Bigg(\bigg(2-z_1\bigg) + \bigg(\dfrac{z_1}{2-z_1}\bigg)\Bigg),$$ from which we obtain $$\dfrac{2(p-1)}{p} < z_1=I(m^2) < 2$$ using WolframAlpha.

Here is my inquiry:

QUESTION: In this closely related MSE question, we were able to derive the improved lower bound $$\frac{2(p-1)}{p}+\frac{1}{pm^2}<I(m^2).$$ Can we similarly derive an improved upper bound for $I(m^2)$, that is hopefully better than $$I(m^2) \leq \frac{2p}{p+1}?$$ If we cannot, then can you explain why?

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MY ATTEMPT

Consider the quantity $$\bigg(\dfrac{D(m^2)}{m^2} - \dfrac{2}{p+1}\bigg)\bigg(\dfrac{\sigma(m^2)}{D(m^2)} - \dfrac{2}{p+1}\bigg).$$ This quantity is nonnegative. Thus, we obtain $$I(m^2) + \bigg(\dfrac{2}{p+1}\bigg)^2 \geq \dfrac{2}{p+1}\bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\bigg) = \dfrac{2}{p+1}\Bigg(\bigg(2-I(m^2)\bigg) + \bigg(\dfrac{I(m^2)}{2-I(m^2)}\bigg)\Bigg).$$ Now, let $z_2 = I(m^2)$. Then we have the inequality $$z_2 + \bigg(\dfrac{2}{p+1}\bigg)^2 \geq \dfrac{2}{p+1}\Bigg(\bigg(2-z_2\bigg) + \bigg(\dfrac{z_2}{2-z_2}\bigg)\Bigg),$$ from which we obtain $$\dfrac{4}{p+3} \leq z_2=I(m^2) \leq \dfrac{2p}{p+1},$$ using WolframAlpha, which does not improve on the previous known bounds for $I(m^2)$.

Answer 1

This is a partial answer.

(I merely wanted to collect, in one place, some recent thoughts that occurred to me, while I was pondering more on the problem in the original post. As you can see, this is too long to fit in the Comments section.)

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Consider the quantity $$\bigg(\dfrac{D(m^2)}{m^2} - p\bigg)\bigg(\dfrac{\sigma(m^2)}{D(m^2)} - p\bigg).$$ This quantity is nonnegative. Consequently, we obtain $$I(m^2) + p^2 \geq p\Bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\Bigg).$$ But it can be proven that the equations $$\dfrac{D(m^2)}{m^2} = \dfrac{2\sigma(p^{k-1})}{\sigma(p^k)}$$ and $$\dfrac{\sigma(m^2)}{D(m^2)} = \dfrac{p^k}{\sigma(p^{k-1})}$$ hold in general. These equations can be rewritten as $$\dfrac{D(m^2)}{m^2} = \dfrac{2\bigg(\dfrac{p^k - 1}{p-1}\bigg)}{\dfrac{p^{k+1} - 1}{p-1}} = \dfrac{2(p^k - 1)}{p^{k+1} - 1}$$ and $$\dfrac{\sigma(m^2)}{D(m^2)} = \dfrac{p^k (p - 1)}{p^k - 1}$$ so that $$\dfrac{D(m^2)}{m^2}+\dfrac{\sigma(m^2)}{D(m^2)}=\dfrac{2(p^k - 1)}{p^{k+1} - 1}+\dfrac{p^k (p - 1)}{p^k - 1}$$ which WolframAlpha expands to $$\dfrac{D(m^2)}{m^2}+\dfrac{\sigma(m^2)}{D(m^2)}=\dfrac{p^{2k+2} - p^{2k+1} + 2p^{2k} - p^{k+1} - 3p^k + 2}{(p^{k+1} - 1)(p^k - 1)} := f(p,k).$$

But $$\dfrac{\partial f}{\partial k}=-\dfrac{p^k (p-1) \Bigg(p^k \bigg((p^2 - 2)p^k - 2(p - 2)\bigg) - 1\Bigg) \log(p)}{(p^{k+1} - 1)^2 (p^k - 1)^2} < 0$$ which means that $f(p,k)$ is a decreasing function of $k$. This implies that $$\dfrac{D(m^2)}{m^2}+\dfrac{\sigma(m^2)}{D(m^2)}=f(p,k) \geq \lim_{k \rightarrow \infty}{f(p,k)}=\lim_{k \rightarrow \infty}\Bigg({\Bigg(\dfrac{2\bigg(1-\dfrac{1}{p^k}\bigg)}{p-\dfrac{1}{p^k}}\Bigg)+\Bigg(\dfrac{p-1}{1-\dfrac{1}{p^k}}\Bigg)}\Bigg).$$ Since $p$ is a prime satisfying $p \equiv 1 \pmod 4$, then $p \geq 5$, so that $$\lim_{k \rightarrow \infty}\Bigg({\Bigg(\dfrac{2\bigg(1-\dfrac{1}{p^k}\bigg)}{p-\dfrac{1}{p^k}}\Bigg)+\Bigg(\dfrac{p-1}{1-\dfrac{1}{p^k}}\Bigg)}\Bigg)=\dfrac{2(1-0)}{p-0}+\dfrac{p-1}{1-0}=\dfrac{2}{p}+(p-1).$$

Hence, we obtain $$I(m^2) + p^2 \geq p\Bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\Bigg) \geq p\Bigg(\dfrac{2}{p}+(p-1)\Bigg),$$ which results in the trivial lower bound $$I(m^2) \geq 2 - p.$$

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Again, consider the quantity $$\bigg(\dfrac{D(m^2)}{m^2} - \dfrac{2}{p+1}\bigg)\bigg(\dfrac{\sigma(m^2)}{D(m^2)} - \dfrac{2}{p+1}\bigg).$$ This quantity is nonnegative. Consequently, we obtain $$I(m^2) + \bigg(\dfrac{2}{p+1}\bigg)^2 \geq \dfrac{2}{p+1}\Bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\Bigg).$$ Proceeding similarly as above, we get $$\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)} \geq \dfrac{2}{p}+(p-1)$$ so that $$I(m^2) + \bigg(\dfrac{2}{p+1}\bigg)^2 \geq \dfrac{2}{p+1}\Bigg(\dfrac{2}{p}+(p-1)\Bigg)$$ from which we finally obtain $$I(m^2) \geq \dfrac{2(p^3 - p + 2)}{p(p+1)^2}$$ which does not improve on the previous known lower bound $$I(m^2) > \dfrac{2(p-1)}{p}.$$

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Lastly, let us (again) consider the quantity $$\bigg(\dfrac{D(m^2)}{m^2} - \dfrac{2}{p}\bigg)\bigg(\dfrac{\sigma(m^2)}{D(m^2)} - \dfrac{2}{p}\bigg).$$ This quantity is negative. Consequently, we obtain $$I(m^2) + \bigg(\dfrac{2}{p}\bigg)^2 < \dfrac{2}{p}\Bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\Bigg).$$ Proceeding similarly as above, since $$\dfrac{D(m^2)}{m^2}+\dfrac{\sigma(m^2)}{D(m^2)}=\dfrac{p^{2k+2} - p^{2k+1} + 2p^{2k} - p^{k+1} - 3p^k + 2}{(p^{k+1} - 1)(p^k - 1)} = f(p,k)$$ is a decreasing function of $k$, and since $k$ is a positive integer satisfying $k \equiv 1 \pmod 4$, then $k \geq 1$, from which it follows that $$\dfrac{D(m^2)}{m^2}+\dfrac{\sigma(m^2)}{D(m^2)}=f(p,k) \leq f(p,1)=\dfrac{2}{p+1}+p$$ so that we obtain $$I(m^2) + \bigg(\dfrac{2}{p}\bigg)^2 < \dfrac{2}{p}\Bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\Bigg) \leq \dfrac{2}{p}\Bigg(\dfrac{2}{p+1}+p\Bigg)$$ from which we finally get $$I(m^2) < \dfrac{2(p-1)(p^2 + 2p + 2)}{p^2 (p + 1)}$$ which does not improve on the known upper bound $$I(m^2) \leq \dfrac{2p}{p+1}.$$

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Added: October 13, 2021 - 9:09 AM (Manila time)

Consider the quantity $$\bigg(\dfrac{D(m^2)}{m^2} - (p-1)\bigg)\bigg(\dfrac{\sigma(m^2)}{D(m^2)} - (p-1)\bigg).$$ This quantity is negative. Proceeding similarly as above, we obtain $$I(m^2) + \bigg(p-1\bigg)^2 < \bigg(p-1\bigg)\Bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\Bigg) \leq \bigg(p-1\bigg)\Bigg(\dfrac{2}{p+1}+p\Bigg)$$ which implies that $$I(m^2) < \dfrac{(p+3)(p-1)}{p+1}.$$ This resulting upper bound is trivial, as it does not improve on the known upper bound $$I(m^2) \leq \dfrac{2p}{p+1}.$$

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Alas, this is where I get stuck!

Answer 2

This is an experimental attempt - my apologies for any silly mistakes.

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Since we have $$I(m^2) \leq \dfrac{2p}{p+1}$$ and because $p^k m^2$ is perfect, then we have $$I(p^k)I(m^2) = I(p^k m^2) = 2$$ where we have used the fact that the abundancy index function is multiplicative.

Hence, we obtain (via iteration) $$\text{First iteration: } I(m^2) \leq I(p^k)I(m^2)\cdot\dfrac{p}{p+1} \leq 2I(p^k)\cdot\dfrac{p^2}{(p+1)^2} = \dfrac{2p^2\Bigg(p^{k+1} - 1\Bigg)}{p^k (p-1)(p+1)^2}$$ $$\text{Second iteration: } I(m^2) \leq 2I(p^k)\cdot\dfrac{p^2}{(p+1)^2}$$ $$= \bigg(I(p^k)\bigg)^2 {I(m^2)} \cdot\dfrac{p^2}{(p+1)^2} \leq 2\bigg(I(p^k)\bigg)^2 \cdot\dfrac{p^3}{(p+1)^3}$$ $$\ldots$$ $$\ldots$$ $$\ldots$$ by recursively replacing $2$ with $I(p^k)I(m^2)$ and then bounding $I(m^2)$ from above by $2p/(p+1)$.

Note that, at the $n^{\text{th}}$ iteration, we have the inequality $$I(m^2) \leq 2\bigg(I(p^k)\bigg)^n \cdot \dfrac{p^{n+1}}{(p+1)^{n+1}}.$$ Repeating the process ad infinitum, we get $$I(m^2) \leq \lim_{n \rightarrow \infty}{\Bigg(2\bigg(I(p^k)\bigg)^n \cdot \dfrac{p^{n+1}}{(p+1)^{n+1}}\Bigg)} = \lim_{n \rightarrow \infty}{\Bigg(\dfrac{2\bigg(I(p^k)\bigg)^n}{\dfrac{(p+1)^{n+1}}{p^{n+1}}}\Bigg)},$$ a limit which is of the indeterminate form $$\dfrac{+\infty}{+\infty}.$$

Hence, we may apply L'Hôpital's rule, by separately considering $$h(p) = \Bigg(\dfrac{2\bigg(I(p^k)\bigg)^n}{\dfrac{(p+1)^{n+1}}{p^{n+1}}}\Bigg)$$ as a function of $p$, and $$j(k) = \Bigg(\dfrac{2\bigg(I(p^k)\bigg)^n}{\dfrac{(p+1)^{n+1}}{p^{n+1}}}\Bigg)$$ as a function of $k$.

(Also, note that there is "currently no L'Hôpital's rule for multiple variable limits". The closest thing that I could find on arXiv is this preprint by Gary R. Lawlor.)

I will stop here for the time being.

Update: (October 13, 2021 - 11:37 AM) - Manila time Since the denominator of $j(k)$ is strictly a function of $p$ and $n$, then there is no need to consider $j(k)$.

Working on $h(p)$ now, will post an update in a bit.

Update: (October 13, 2021 - 12:27 PM) - Manila time Cross-posted this answer as a question to MO, since the computations involved are somewhat tedious and complicated.