Does the following lower bound improve on $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number? - Part II

Question

Preamble: This question is an offshoot of these earlier posts: (post1), (post2).

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 $$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$

In (post2), I derived the following inequality: $$2qn^2 - q\sigma(n^2) < 2n^2 - 1.$$

Using my method, I then get the bounds $$I(q^k) < \dfrac{q}{q-1} < \dfrac{2(q-1)}{q} + \dfrac{1}{qn^2} < I(n^2)$$ which implies that the product $$(I(q^k) - x)(I(n^2) - x) < 0$$ is negative, where $$x = \dfrac{2(q-1)}{q} + \dfrac{1}{qn^2}.$$ Upon some algebra, I get $$I(q^k) + I(n^2) > \dfrac{3q^2 - 4q + 2}{q(q - 1)} - \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2}.$$ But $$- \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2} = \dfrac{qn^2 (q - 4) + q + 2n^2 - 1}{qn^2 (q - 1)(2n^2 (q - 1) + 1)} > 0,$$ since $q$ is a prime satisfying $q \equiv 1 \pmod 4$ implies that $q \geq 5$.

So it does appear that the inequality $$I(q^k) + I(n^2) > \dfrac{3q^2 - 4q + 2}{q(q - 1)} - \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2}$$ is unconditionally true, which would mean that the new lower bound for $I(q^k) + I(n^2)$ improves on the old. Note that we were able 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 $k \leq a$?

Answer 1

Let $f(k):=I(q^k)+\dfrac{2}{I(q^k)}$ and $$g(q,n):=\dfrac{3q^2 - 4q + 2}{q(q - 1)} - \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2}.$$

One has $$f'(k)=\frac{-( q^{2 k + 2}- 4 q^{2 k + 1}+2 q^{k + 1} + 2 q^{2 k} - 1) \log(q)}{(q - 1)q^k (q^{k + 1} - 1)^2}\lt 0$$ so it follows that $f(k)$ is strictly decreasing with $$\dfrac{3q^2 - 4q + 2}{q(q - 1)}=\lim_{k\to\infty}f(k)\lt f(k)\le f(1)=\frac{3q^2+2q+1}{q(q+1)}.$$

You have already known that $f(k)\gt g(q,n)\gt \lim_{k\to\infty}f(k)$ and that $f(1)\gt g(q,n)$ which is equivalent to $2n^2\gt q+1$.

It follows that there exists $K$ satisfying $1\leqslant k\lt K$ and $f(K)=g(q,n)$.

Answer 2

This post complements mathlove's answer here, and thereby computes an explicit upper bound for $K$ in terms of $q$ and $n$.

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So here we go: We require $f(K) = g(q,n)$.

But we have the partial fraction decompositions $$f(K) = \frac{3q^2 - 4q + 2}{q(q - 1)} + \frac{2(q - 1)}{q(q^{K+1} - 1)} - \frac{1}{{q^K}(q - 1)}$$ and $$g(q,n) = \frac{3q^2 - 4q + 2}{q(q - 1)} - \frac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \frac{1}{qn^2},$$ as computed by WolframAlpha.

Rearranging terms from the equation $f(K) = g(q,n)$ then gives $$\frac{2(q - 1)}{q(q^{K+1} - 1)} - \frac{1}{qn^2} = \frac{1}{{q^K}(q - 1)} - \frac{q}{(q - 1)(2qn^2 - 2n^2 + 1)}.$$

After some algebraic simplifications, we obtain $$\frac{2n^2 (q - 1) - (q^{K+1} - 1)}{qn^2 (q^{K+1} - 1)} = \frac{2qn^2 - 2n^2 + 1 - q^{K+1}}{{q^K} (q - 1)(2qn^2 - 2n^2 + 1)},$$ from which we finally get $$\frac{2qn^2 - 2n^2 + 1 - q^{K+1}}{qn^2 (q^{K+1} - 1)} = \frac{2qn^2 - 2n^2 + 1 - q^{K+1}}{{q^K} (q - 1)(2qn^2 - 2n^2 + 1)}.$$

Suppose to the contrary that $$2qn^2 - 2n^2 + 1 - q^{K+1} \neq 0.$$ Then we may cancel $2qn^2 - 2n^2 + 1 - q^{K+1}$ in the numerator of both sides of the equation, to get $${q^K} (q - 1)(2qn^2 - 2n^2 + 1) = qn^2 (q^{K+1} - 1).$$ This may be rewritten as $$2 - \frac{2}{q} + \frac{1}{qn^2} = \frac{2qn^2 - 2n^2 + 1}{qn^2} = \frac{q^{K+1} - 1}{q^K (q - 1)}.$$ But we know of the estimates $$\frac{2(q - 1)}{q} < 2 - \frac{2}{q} + \frac{1}{qn^2} = \frac{q^{K+1} - 1}{q^K (q - 1)} < \frac{q}{q - 1}.$$ These estimates imply that $$\sqrt{2} < \frac{q}{q - 1},$$ contradicting $$\frac{q}{q - 1} \leq \frac{5}{4},$$ since $q$ is the special prime satisfying $q \equiv 1 \pmod 4$ implies that $q \geq 5$.

The contradiction thus obtained means that our assumption that $$2qn^2 - 2n^2 + 1 - q^{K+1} \neq 0$$ is untenable. This implies that $$2qn^2 - 2n^2 + 1 - q^{K+1} = 0,$$ from which we obtain $$2n^2 (q - 1) = q^{K+1} - 1$$ $$2n^2 = \frac{q^{K+1} - 1}{q - 1}$$

This implies that $$q^K < 2n^2$$ $$K \log{q} < \log{2} + 2\log{n}$$ Finally, we get the upper bound $$K < \log_q{2} + 2\log_q{n}.$$

Answer 3

This post generalizes this answer, since the earlier one is getting too long already.

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Abbreviate a (strict) lower bound for the quantity $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))$$ by $\rho$.

Using my method, we then get the bounds $$I(q^k) < \dfrac{q}{q-1} < \dfrac{2(q-1)}{q} + \dfrac{\rho}{qn^2} < I(n^2)$$ which implies that the product $$(I(q^k) - y)(I(n^2) - y) < 0$$ is negative, where $$y = \dfrac{2(q-1)}{q} + \dfrac{\rho}{qn^2}.$$ After some careful algebraic simplifications, I get $$I(q^k) + I(n^2) > \frac{2qn^2}{2qn^2 - 2n^2 + \rho} + \frac{2qn^2 - 2n^2 + \rho}{qn^2}$$ which has the partial fraction decomposition $$\frac{2qn^2}{2qn^2 - 2n^2 + \rho} + \frac{2qn^2 - 2n^2 + \rho}{qn^2} = \frac{3q^2 - 4q + 2}{q(q - 1)} - \frac{\rho q}{(q - 1)(2qn^2 - 2n^2 + \rho)} + \frac{\rho}{qn^2}.$$ Therefore, $$I(q^k) + I(n^2) > \dfrac{3q^2 - 4q + 2}{q(q - 1)} - \dfrac{\rho q}{(q - 1)(2qn^2 - 2n^2 + \rho)} + \dfrac{\rho}{qn^2}.$$

Since it is known that $\rho > 1$ holds, and that $q \geq 5$, then we also know that $$-\frac{\rho q}{(q - 1)(2qn^2 - 2n^2 + \rho)} + \frac{\rho}{qn^2} = \frac{\rho \Bigg(qn^2 (q - 4) + \rho(q - 1) + 2n^2\Bigg)}{qn^2 (q - 1)(2n^2 (q - 1) + \rho)} > 0.$$

This means that the new lower bound $$I(q^k) + I(n^2) > h(q,n)$$ where $$h(q,n) = \frac{2qn^2}{2qn^2 - 2n^2 + \rho} + \frac{2qn^2 - 2n^2 + \rho}{qn^2}$$ improves on the old (and trivial) lower bound $$I(q^k) + I(n^2) > \frac{3q^2 - 4q + 2}{q(q - 1)}.$$ Note that $h(q,n)$ does not contain $k$.

By mathlove's result, we know that there exists a number $K'$ such that $1 \leq k < K'$.

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We now compute an explicit upper bound for $K'$, in terms of $q$, $n$, and $\rho$.

So here we go: We require $f(K') = h(q,n)$.

But we have the partial fraction decompositions $$f(K') = \frac{3q^2 - 4q + 2}{q(q - 1)} + \frac{2(q - 1)}{q(q^{K' + 1} - 1)} - \frac{1}{q^{K'}(q - 1)}$$ and $$h(q,n) = \frac{3q^2 - 4q + 2}{q(q - 1)} - \frac{\rho q}{(q - 1)(2qn^2 - 2n^2 + \rho)} + \frac{\rho}{qn^2}.$$

Equating and rearranging as before, we obtain $$\frac{2(q - 1)}{q(q^{K' + 1} - 1)} - \frac{\rho}{qn^2} = \frac{1}{q^{K'}(q - 1)} - \frac{\rho q}{(q - 1)(2qn^2 - 2n^2 + \rho)}.$$

After some algebraic simplifications, we get $$\frac{2n^2 (q - 1) - \rho\bigg(q^{K' + 1} - 1\bigg)}{qn^2 \bigg(q^{K' + 1} - 1\bigg)} = \frac{2n^2 (q - 1) + \rho - \rho q^{K' + 1}}{{q^{K'}}(q - 1)(2qn^2 - 2n^2 + \rho)}.$$

Proceeding similarly as in the other answer, suppose to the contrary that $$2n^2 (q - 1) - \rho\bigg(q^{K' + 1} - 1\bigg) \neq 0.$$

Then we can cancel the numerator of both sides of the equation, since $$2n^2 (q - 1) - \rho\bigg(q^{K' + 1} - 1\bigg) = 2n^2 (q - 1) + \rho - \rho q^{K' + 1}.$$

We thus obtain $$qn^2 \bigg(q^{K' + 1} - 1\bigg) = {q^{K'}}(q - 1)(2qn^2 - 2n^2 + \rho)$$ which can be rewritten as $$\frac{q^{K' + 1} - 1}{{q^{K'}}(q - 1)} = \frac{2qn^2 - 2n^2 + \rho}{qn^2}.$$

But, as before, we have the estimates $$\frac{2(q - 1)}{q} < \frac{2qn^2 - 2n^2 + \rho}{qn^2}$$ (since $\rho$ is positive), and $$\frac{q^{K' + 1} - 1}{{q^{K'}}(q - 1)} < \frac{q}{q - 1},$$ which (again) implies that $$\sqrt{2} < \frac{q}{q - 1},$$ contradicting $$\frac{q}{q - 1} \leq \frac{5}{4}$$ since $q$ is the special prime satisfying $q \equiv 1 \pmod 4$ implies that $q \geq 5$.

The contradiction thus obtained means that our assumption that $$2n^2 (q - 1) - \rho\bigg(q^{K' + 1} - 1\bigg) \neq 0$$ is untenable. This implies that $$\frac{2n^2}{\rho} = \frac{q^{K' + 1} - 1}{q - 1}.$$

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We finally obtain the inequality $$q^{K'} < \frac{2n^2}{\rho}$$ which implies that $$K' < \log_q {2} + 2\log_q {n} - \log_q {\rho}.$$

MINOR CAVEAT: As correctly pointed out by mathlove, $K'$ (and therefore also $K$ in the other answer) may not be an integer. Hence, we may not write $\sigma(q^{K'})$ or $I(q^{K'})$ (nor $\sigma(q^K)$ and $I(q^{K})$ in the other answer).

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Per this answer to a tangentially related question, the best currently known lower bound for $\sigma(n^2)/q^k$ is $$\frac{\sigma(n^2)}{q^k} \geq {3^3} \times {5^3} = 3375.$$