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Let $$\sigma(x) = \sum_{l \mid x}{l}.$$ That is, let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$ (the set of natural numbers or positive integers). Set $$D(x) := 2x - \sigma(x)$$ to be the deficiency of the number $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

If a number $n$ satisfies $\sigma(n)=2n$, then $n$ is said to be perfect. For example, the number $6$ is perfect since the divisors of $6$ are $1$, $2$, $3$ and $6$, and the sum of the divisors of $6$ is $$\sigma(6)=1+2+3+6=12=2\cdot{6}.$$

Currently, only $50$ even perfect numbers are known, and they are all of the form $(2^q - 1){2^{q-1}}$ as shown by the Euclid-Euler Theorem. On the other hand, no odd perfect numbers have been found. It is known that an odd perfect number must be bigger than ${10}^{1500}$, by a fairly recent result of Ochem and Rao. Lastly, Euler proved that an odd perfect number must have the form $$n=p^k m^2$$ where $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Now, by a mild brute-force computation, it is possible to prove that: $$D(p^k)D(m^2)=2(\sigma(p^k) - p^k)(\sigma(m^2) - m^2).$$

Dividing both sides by $n=p^k m^2$, we obtain $$\frac{D(p^k)}{p^k}\cdot\frac{D(m^2)}{m^2}=2(I(p^k) - 1)(I(m^2) - 1)=(2 - I(p^k))(2 - I(m^2))=2(3 - (I(p^k)+I(m^2))).$$

Using the estimate $$D(x) \leq \phi(x)$$ where $\phi(x)$ is the Euler totient of $x$, and the fact that the Euler totient is multiplicative, we get $$2(3 - (I(p^k)+I(m^2)))=\frac{D(p^k)}{p^k}\cdot\frac{D(m^2)}{m^2} \leq \frac{\phi(p^k)}{p^k}\cdot\frac{\phi(m^2)}{m^2} = \frac{\phi(n)}{n} < \frac{1}{2},$$ where the last inequality is due to Advanced Problem H-661, On Odd Perfect Numbers, Proposed by J. L´opez Gonz´alez, Madrid, Spain and F. Luca, Mexico (Vol. 45, No. 4, November 2007), Fibonacci Quarterly. This implies that $$I(p^k) + I(m^2) > \frac{11}{4} = 2.75,$$ which is trivial compared to the lower bound $$I(p^k) + I(m^2) \gneq 2\sqrt{I(p^k)I(m^2)} = 2\sqrt{2} \approx 2.828\cdots,$$ where inequality is due to the fact that $I(p^k) = I(m^2) = \sqrt{I(p^k)I(m^2)} = \sqrt{2}$ is untenable.

Here is my question:

Does anybody here have any ideas on how to get good bounds for the expression $$\frac{D(p^k)}{p^k}\cdot\frac{D(m^2)}{m^2}=2(I(p^k) - 1)(I(m^2) - 1)=(2 - I(p^k))(2 - I(m^2))=2(3 - (I(p^k)+I(m^2)))?$$

Currently, I can get $$\frac{2(p - 1)}{p(p + 1)} \leq \bigg(\frac{D(p^k)}{p^k}\bigg)\bigg(\frac{D(m^2)}{m^2}\bigg) < \frac{2(p - 2)}{p(p - 1)}.$$ Equality holds if and only if $k=1$. However, this approach does not seem to lend itself well to numerical approximations/bounding, as we only have the lower bound $p \geq 5$ but not an upper bound.

Reference

The abundancy index of divisors of odd perfect numbers – Part III, Notes on Number Theory and Discrete Mathematics, Volume 23, 2017, Number 3, Pages 53—59

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  • $\begingroup$ From the upper bound for $(D(p^k)/p^k)\cdot(D(m^2).m^2)$ in terms of $p$, I can easily get $$\frac{2(p-2)}{p(p-1)} < \frac{2}{p} \leq \frac{2}{5},$$ from which we get the lower bound $$I(p^k) + I(m^2) > 3 - \frac{1}{5} = \frac{14}{5} = 2.8,$$ which is still trivial compared to the lower bound obtained when using the Arithmetic Mean-Geometric Mean Inequality. $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 7 '18 at 6:25
  • $\begingroup$ When the Euler prime $p=5$, then I get the bounds $$\frac{57}{20} < I(p^k) + I(m^2) \leq \frac{43}{15}.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 7 '18 at 11:06

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