Let
$$\sigma(x) = \sum_{e \mid x}{e}$$
denote the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, and the deficiency of $x$ by $D(x)=2x-\sigma(x)$. A positive integer $N$ is said to be deficient-perfect if $D(N) \mid N$.
Here is my question:
Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers $N > 1$?
$$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}$$
(Note that the inequality
$$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}$$
is true if and only if $N$ is deficient.)
References
Posted Answer #1
ILLUSTRATING VIA A TOY EXAMPLE
Let $M$ be an odd perfect number given in the so-called Eulerian form
$$M = p^k m^2$$
(i.e. $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$).
It is known that the non-Euler part $m^2$ is deficient-perfect if and only if the Descartes-Frenicle-Sorli conjecture that $k=1$ holds. (See this paper for a proof of this fact.)
So, suppose that $k=1$. Then $m^2$ is deficient-perfect.
In particular, $m^2$ is deficient, so that the criterion in this paper applies.
We have
$$\frac{2m^2}{m^2 + D(m^2)} < I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)}.$$
Under the hypothesis that $k=1$, $m^2$ is deficient-perfect, with deficiency
$$D(m^2) = \frac{m^2}{(p+1)/2}.$$
We also have
$$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p+1}.$$
Putting these all together, we have
$$\frac{m^2}{D(m^2)} = \frac{p+1}{2}$$
$$\frac{2p}{p+1} = I(m^2) > \frac{2m^2}{m^2 + D(m^2)} = \frac{2\bigg(\frac{m^2}{D(m^2)}\bigg)}{\frac{m^2}{D(m^2)} + 1} = \frac{2\bigg(\frac{p+1}{2}\bigg)}{\bigg(\frac{p+1}{2}\bigg) + 1} = \frac{p+1}{\frac{p+3}{2}} = \frac{2(p+1)}{p+3}$$
which implies that
$$p^2 + 3p = p(p+3) > (p+1)^2 = p^2 + 2p + 1$$
$$p > 1$$
(This last inequality is trivial as $p$ is prime with $p \equiv 1 \pmod 4$ implies that $p \geq 5$.)
$$\frac{2p}{p+1} = I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)} = \frac{2\bigg(\frac{m^2}{D(m^2)}\bigg) + 1}{\frac{m^2}{D(m^2)} + 1} = \frac{2\bigg(\frac{p+1}{2}\bigg) + 1}{\bigg(\frac{p+1}{2}\bigg) + 1} = \frac{p+2}{\frac{p+3}{2}} = \frac{2(p+2)}{p+3}$$
which implies that
$$p^2 + 3p = p(p+3) < (p+1)(p+2) = p^2 + 3p + 2$$
$$0 < 2.$$
This example illustrates my interest in improvements to the bounds in terms of the abundancy index and deficiency functions of $N$, when $N > 1$ is deficient-perfect.
Posted Answer #2
Suppose that $N > 1$ is deficient-perfect. Since $N$ is deficient, then
$$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}.$$
I think that, since $D(N) \mid N$ when $N$ is deficient-perfect, then $N/D(N)$ is an integer, so that we have (since $\frac{N}{D(N)} \mid N$)
$$I\bigg(\frac{N}{D(N)}\bigg) \leq I(N) < \frac{2N + D(N)}{N + D(N)} = \frac{2\bigg(\frac{N}{D(N)}\bigg) + 1}{\bigg(\frac{N}{D(N)}\bigg) + 1}.$$
CLAIM
$$\frac{2\bigg(\frac{N}{D(N)}\bigg)}{\bigg(\frac{N}{D(N)}\bigg) + 1} < I\bigg(\frac{N}{D(N)}\bigg)$$
This claim, if true, would prove that all deficient-perfect numbers $N$ correspond to almost perfect numbers $N/D(N)$.
Added April 18 2019 (6:13 PM - Manila time)
The claim is false. A counterexample is given by
$$N = \bigg({3}\cdot{7}\cdot{11}\cdot{13}\bigg)^2.$$
Added April 18 2019 (6:17 PM - Manila time)
It appears that the claim is true when $D(N)=1$.
