Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the abundancy index
$$I(x)=\frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the classical sum of divisors of $x$.
Since $q$ is prime, we have the bounds
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$
which implies, since $N$ is perfect, that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
By considering the negative product
$$\bigg(I(q^k) – \frac{2(q-1)}{q}\bigg)\bigg(I(n^2) – \frac{2(q-1)}{q}\bigg) < 0,$$
since we obviously have
$$\frac{q}{q-1} < \frac{2(q-1)}{q},$$
then after some routine algebraic manipulations, we arrive at the lower bound
$$I(q^k) + I(n^2) > 3 – \frac{q-2}{q(q-1)} = \frac{3q^2 – 4q + 2}{q(q – 1)}.$$
Now, a recent MO post improves on the lower bound for $I(n^2)$, as follows:
$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$
Added February 3, 2021 – 8:10 PM (Manila time) Here is a quick way to show the improved lower bound for $I(n^2)$. We have
$$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q – 1)}{q^{k+1} – 1}=\frac{2q^{k+1} (q – 1)}{q(q^{k+1} – 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} – 1}\bigg)$$
$$=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{(q^{k+1} – 1) + 1}{q^{k+1} – 1}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} – 1}\bigg)$$
$$>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)$$
since $q^{k+1} > q^{k+1} – 1$. But, of course, we obtain
$$I(n^2)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg).$$
QED.
Repeating the same procedure as above, we have the negative product
$$\Bigg(I(q^k) – \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg)\Bigg(I(n^2) – \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg) < 0.$$
This implies, after some algebraic manipulations, that
$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q – 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}.$$
But WolframAlpha says that the partial fraction decomposition of the new lower bound is given by
$$\frac{q^{k+2}}{(q – 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}} = \frac{3q^2 – 4q + 2}{q(q – 1)} + \frac{2(q – 1)}{q^{k+2}} – \frac{q}{(q – 1)(q^{k+1} + 1)}.$$
So essentially, my question boils down to:
QUESTION: Is it possible to produce an unconditional proof (that is, for all $k \geq 1$ and for all special primes $q \geq 5$) for the following inequality?
$$\frac{2(q – 1)}{q^{k+2}} > \frac{q}{(q – 1)(q^{k+1} + 1)}$$
MY ATTEMPT
I tried to ask WolframAlpha for a plot of the above inequality, it gave me the following GIF image:
So it does appear that the inequality is unconditionally true, which would mean that the new lower bound for $I(q^k) + I(n^2)$ improves on the old. Is it possible to prove this analytically?
And lastly: Based on this answer to a closely related question, since we appear to have obtained an improved lower bound for $I(q^k) + I(n^2)$, can we then say that there is indeed an integer $a$ such that $k \leq a$?
Best Answer
I think that we cannot.
In the following, I'll explain the reason why I think that we cannot.
In the answer, I wrote
"If you get an improved lower bound for $f(k)$, then you can say that there is an integer $a$ such that $k\le a$ "
where $f(k)=I(q^k) + I(n^2)$.
Also, in a comment below the answer, I wrote
"if you get an improved lower bound $g(q)$, then from the above fact, we see that there is only one $k=k_0$ such that $f(k)=g(q)$. Then, we have $k\le \lfloor k_0\rfloor$ "
where "the above fact" is the fact that $f(k)$ is decreasing.
In short, we can say that there is an integer $a$ such that $k\le a$ if we get a function $g(q)$ (which is a fuction only on $q$) such that $$I(q^k)+I(n^2)\ge g(q)\gt 3 - \frac{q-2}{q(q - 1)}$$
Now, let us see what you've got.
Let $$G(k,q)=\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}$$
You have $$I(q^k) + I(n^2) > G(k,q)\gt 3 - \frac{q-2}{q(q - 1)}$$ which is correct since
$$I(q^k) + I(n^2)-G(k,q)=\frac{ (q- 4)q^{k+2}+2 q^{k + 1} + (3 q - 4 )q + 2}{q^{k+2}(q - 1) (q^{k + 1} - 1) (q^{k + 1} + 1)}\gt 0$$
$$G(k,q)-\bigg(3 - \frac{q-2}{q(q - 1)}\bigg)=\frac{ (q - 4) q^{k + 2} +2 q^{k + 1} + 2( q - 2 )q + 2}{q^{k+2}(q - 1) (q^{k + 1} + 1)}\gt 0$$
Also, we have $$\frac{\partial G(k,q)}{\partial k}=-\frac{ (q- 4) q^{2 k + 3}+ 2 q^{2 k + 2}+ 4( q- 2)q^{k + 2} +4 q^{k + 1} + 2( q - 2) q + 2}{q^{k+2}(q - 1) (q^{k + 1} + 1)^2}\ln q\lt 0$$ $$\lim_{k\to\infty}\bigg(I(q^k) + I(n^2) \bigg)=\lim_{k\to\infty}G(k,q)= 3 - \frac{q-2}{q(q - 1)}$$
These imply that you don't get a function $g(q)$ (which is a fuction only on $q$) such that $$I(q^k)+I(n^2)\ge g(q)\gt 3 - \frac{q-2}{q(q - 1)}$$
So, I think that you cannot say that there is an integer $a$ such that $k\le a$.
Added :
I'm going to add another explanation with graphs.
Let $$f(k)=I(q^k) + I(n^2),\qquad G(k)=\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}},$$ $$Q=3 - \dfrac{q-2}{q(q - 1)}$$ where $q,Q$ are seen as constants.
Since we see that $f(k)\gt G(k)\gt Q$ holds for all $k,q$ and that both $y=f(k)$ and $y=G(k)$ are decreasing with $\displaystyle\lim_{k\to\infty}f(k)=\displaystyle\lim_{k\to\infty}G(k)=Q$, we see that their graphs are as follows :
Now, if you take a fixed positive integer $k=k'$, then we have the following :
This explains why you cannot say that $$f(k)\gt G(k′)$$ holds for all $k$.
Added 2 :
I'm going to add more explanations with steps :
(1) $f(k)\gt G(k)\gt Q$ holds for all $k,q$.
(2) Both $y=f(k)$ and $y=G(k)$ are strictly decreasing.
(3) $\displaystyle\lim_{k\to\infty}f(k)=\displaystyle\lim_{k\to\infty}G(k)=Q$.
(4) If we let $k=k'$ where $k'$ is a fixed positive number larger than $1$, then $G(k')$ is entirely in terms of $q$ alone.
(5) $Q\lt G(k')\lt f(1)$
(6) It follows from $(2)(3)$ that, for any $Y$ satisfying $Q\lt Y\lt f(1)$, there exists only one $k$ such that $f(k)=Y$.
(7) It follows from $(5)(6)$ that there exists only one $k$ such that $f(k)=G(k')$. Let $k_0$ be this $k$.
(8) For $k\lt k_0$, we have $f(k)\gt G(k')$.
(9) For $k\ge k_0$, we have $f(k)\le G(k')$.
(10) It follows from $(8)(9)$ that $f(k)\gt G(k')$ does not hold for all $k$.
Added 3 :
You seem to think that the following claim is true :
Claim : If there are two functions $f(x),g(x)$ such that $$f(x)\gt g(x)\gt 1$$ holds for all $x\gt 0$, then there is a fixed positive integer $m$ such that $f(x)\gt g(m)$ holds for all $x\gt 0$.
This claim is false. Take $f(x)=\dfrac 1x+1$ and $g(x)=\dfrac{1}{x+1}+1$ for which $f(x)\gt g(x)\gt 1$ holds for all $x\gt 0$, but $f(x)\gt g(m)$ does not hold for $x\ge m+1$.