Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$?

Question

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

Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.

Here is my initial question:

Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ lead to an improvement to the upper bound $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$?

MOTIVATION

Here is a proof that $I(n^2) \leq 2 - \frac{5}{3q}$ holds in general.

Suppose to the contrary that $I(n^2) > 2 - \frac{5}{3q}$ is true.

Note that $$I(n^2) = \frac{2}{I(q^k)} = \frac{2q^k (q - 1)}{q^{k+1} - 1} = 2 - 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg),$$ so that we have $$I(n^2) > 2 - \frac{5}{3q} \iff 2 - 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg) > 2 - \frac{5}{3q} \iff \frac{5}{3q} > 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg) \iff 5q^{k+1} - 5 > 6q^{k+1} - 6q \iff 0 > q^{k+1} - 6q + 5,$$ which then implies that $k=1$. (Otherwise, if $k > 1$, we have $$0 > q^{k+1} - 6q + 5 \geq q^6 - 6q + 5,$$ since $k \equiv 1 \pmod 4$, contradicting $q \geq 5$.) Now, since $k=1$, we get $$0 > q^2 - 6q + 5 = (q - 1)(q - 5),$$ which implies that $1 < q < 5$. This contradicts $q \geq 5$. This concludes the proof.

Now, let $Q = 2 - \frac{5}{3q}$.

Since $I(q^k) < I(n^2)$, then we obtain $$I(q^k) < I(n^2) \leq Q \iff (I(q^k) - Q)(I(n^2) - Q) \geq 0$$ $$\iff 2 + Q^2 = I(q^k)I(n^2) + Q^2 \geq Q(I(q^k) + I(n^2) \iff I(q^k) + I(n^2) \leq \frac{2}{Q} + Q.$$ But $\frac{2}{Q} + Q$ can be rewritten as $$\dfrac{2}{Q} + Q = \dfrac{2}{2 - \dfrac{5}{3q}} + \Bigg(2 - \dfrac{5}{3q}\Bigg) = 3 - \Bigg(\dfrac{5}{3q} - \dfrac{5}{6q-5}\Bigg) = 3 - \dfrac{5(3q - 5)}{3q(6q - 5)} = \dfrac{54q^2 - 60q + 25}{3q(6q - 5)}.$$

Let $$f(q) = \dfrac{54q^2 - 60q + 25}{3q(6q - 5)}.$$ Then the derivative $$f'(q) = \dfrac{5}{3q^2} - \dfrac{30}{(6q - 5)^2}$$ is positive for $q \geq 5$. This means that $f$ is an increasing function of $q$, which implies that $$I(q^k) + I(n^2) \leq \dfrac{2}{Q} + Q < \lim_{q \rightarrow \infty}{f(q)} = 3.$$

FINAL QUESTION

Can we do better? If that is not possible, can you explain why?

Answer 1

Too long to comment :

There is no such upper bound of the form $2−\dfrac{cq+d}{q^2+aq+b}$ better than $2−\dfrac{2}{q+1}$.

Proof :

Suppose that there is $(a,b,c,d)$ such that $$2 - 2\cdot\dfrac{q^k - 1}{q^{k+1} - 1} \leqslant 2 - \dfrac{cq+d}{q^2 + aq + b} \leqslant 2 - \dfrac{2}{q+1}$$

Then, $$\dfrac{2q^k - 2}{q^{k+1} - 1} \geqslant \dfrac{cq+d}{q^2 + aq + b} \geqslant \dfrac{2}{q+1}$$ is equivalent to $$\underbrace{\bigg((2-c)q^2+(2a-d)q+2b\bigg)}_{A}q^k-2q^2+(c-2a)q+d-2b\geqslant 0\tag1$$ $$(c-2)q^2+(c+d-2a)q+d-2b\geqslant 0\tag2$$

Suppose that $2-c\lt 0$. Then, it follows from $(1)$ that $A$ will be negative for a large $q^k$ and so LHS of $(1)$ will be negative. This is a contradiction. So, $2-c\geqslant 0$.

Suppose that $c-2\lt 0$. Then, it follows from $(2)$ that LHS of $(2)$ will be negative for large $q$. This is a contradiction. So, $c-2\geqslant 0$.

Therefore, we have to have $c=2$ for which we have $$\underbrace{\bigg((2a-d)q+2b\bigg)}_{A}q^k-2q^2+(2-2a)q+d-2b\geqslant 0\tag3$$ $$(2+d-2a)q+d-2b\geqslant 0\tag4$$ It is necessary that $(3)$ holds for $k=1$, so we have to have $$(2a-d-2)q^2+(2b+2-2a)q+d-2b\geqslant 0\tag5$$

Suppose that $2+d-2a\lt 0$. Then, LHS of $(4)$ will be negative for a large $q$. This is a contradiction. So, $2+d-2a\geqslant 0$.

Suppose that $2a-d-2\lt 0$. Then, LHS of $(5)$ will be negative for a large $q$. This is a contradcition. So, $2a-d-2\geqslant 0$.

Therefore, we have to have $d=2a-2$ for which we have $$a-1-b\geqslant 0\tag6$$ $$(b+1-a)q+a-1-b\geqslant 0\tag7$$ Suppose that $b+1-a\lt 0$. Then, LHS of $(7)$ will be negative for a large $q$. This is a contradiction. So, $b+1-a\geqslant 0$.

Therefore, from $(7)$, we have to have $b=a-1$.

Now, since $b=a-1,c=2$ and $d=2a-2$, we finally have $$2 - \dfrac{cq+d}{q^2 + aq + b}=2 - \dfrac{2q+2a-2}{q^2 + aq + a-1}=2-\frac{2(q+a-1)}{(q+1)(q+a-1)}=2-\dfrac{2}{q+1}$$

This means that there is no such upper bound of the form $2−\dfrac{cq+d}{q^2+aq+b}$ better than $2−\dfrac{2}{q+1}$.$\quad\blacksquare$

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Similarly, one can prove that there is no such upper bound of the form $2−\dfrac{dq^2+eq+f}{q^3+aq^2+bq+c}$ better than $2−\dfrac{2}{q+1}$.

Answer 2

This is a partial answer.

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Note that $$I(n^2) = \dfrac{2}{I(q^k)} \leq \dfrac{2}{I(q)} = \dfrac{2q}{q+1},$$ since $k \equiv 1 \pmod 4$ implies that $k \geq 1$.

Following exactly the same steps in the original post, we obtain $$I(q^k) < I(n^2) \leq \dfrac{2q}{q+1}.$$

After working out the details, we get $$I(q^k) + I(n^2) \leq 3 - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg).$$

Set $Q' = 3 - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg)$.

Let us compute the difference $Q' - \Bigg(\dfrac{2}{Q} + Q\Bigg)$ and determine its sign for $q \geq 5$. We obtain

$$Q' - \Bigg(\dfrac{2}{Q} + Q\Bigg) = \dfrac{5(3q - 5)}{3q(6q - 5)} - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg) = -\dfrac{(3q - 8)(q - 5)}{3q(q + 1)(6q - 5)} \leq 0,$$

with equality occurring if and only if $q = 5$.

This shows that $Q'$ is a better upper bound for $I(q^k) + I(n^2)$ than $\dfrac{2}{Q} + Q$.

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In other words, $\dfrac{2}{Q} + Q$ is a trivial upper bound for $I(q^k) + I(n^2)$.

Answer 3

In this post, I will attempt to improve on the upper bound $$I(n^2) \leq \dfrac{2q}{q+1}.$$

(Although this attempt is unsuccessful, I am retaining this answer here, mainly for my own benefit.)

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Following mathlove's lead in the comments, I tried to find $a, b, c, d, e, f \in \mathbb{R}$ such that $$2 - 2\cdot\Bigg(\dfrac{q^k - 1}{q^{k+1} - 1}\Bigg) = \frac{2q^k (q - 1)}{q^{k+1} - 1} \leq 2 - \Bigg(\dfrac{dq^2 + eq + f}{q^3 + aq^2 + bq + c}\Bigg) \leq 2 - \dfrac{2}{q+1} = \dfrac{2q}{q+1}.$$

This was very messy. But these inequalities are equivalent to $$\dfrac{2}{q+1} \leq \dfrac{dq^2 + eq + f}{q^3 + aq^2 + bq + c} \leq 2\cdot\Bigg(\dfrac{q^k - 1}{q^{k+1} - 1}\Bigg),$$ and I set $$a = b = e = f = 0,$$ and then noticed that $$(c, d) = (25, 2)$$ seems to work.

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PROOF

Let $$S = 2 - \dfrac{2q^2}{q^3 + 25}.$$

We compute $$S - I(n^2) = 2\cdot\Bigg(\dfrac{q^k - 1}{q^{k+1} - 1}\Bigg) - \dfrac{2q^2}{q^3 + 25} = \dfrac{2(25q^k - q^3 + q^2 - 25)}{(q^3 + 25)(q^{k+1} - 1)}.$$

Set $$f(k) := \dfrac{2(25q^k - q^3 + q^2 - 25)}{(q^3 + 25)(q^{k+1} - 1)}.$$

Differentiating once with respect to $k$, we obtain $$f'(k) = \dfrac{2(q - 1){q^k}\log(q)}{(q^{k+1} - 1)^2},$$ which is positive. This means that $f(k)$ is an increasing function of $k$, which implies that $$f(k) \geq f(1) = -\dfrac{2(q^2 - 25)}{(q + 1)(q^3 + 25)}.$$

This implies that $$I(n^2) \leq S + \dfrac{2(q^2 - 25)}{(q + 1)(q^3 + 25)} = \Bigg(2 - \dfrac{2q^2}{q^3 + 25}\Bigg) + \dfrac{2(q^2 - 25)}{(q + 1)(q^3 + 25)} = \dfrac{2q}{q+1},$$ which seems to show, at least, for these particular values for $$(a, b, c, d, e, f) = (0, 0, 25, 2, 0, 0)$$ that $$I(n^2) \leq \dfrac{2q}{q+1}$$ is best-possible.