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This question is an offshoot of the following answer to a closely related MSE question.

Let $N$ be a deficient-perfect number, i.e. $N$ is a positive integer such that $D(N) \mid N$ where $D(N)=2N-\sigma(N)$ is the deficiency and $\sigma(N)$ is the sum of divisors of $N$, respectively.

It is known that $N$ satisfies the inequality $$\frac{2N}{N + D(N)} < I(N) < \frac{2N+D(N)}{N+D(N)}$$ where $I(N)=\sigma(N)/N$ is the abundancy index of $N$, since $N$ is deficient-perfect implies that $N$ is deficient.

Consequently, since $N$ is deficient-perfect, then $N/D(N)$ is an integer, and since $N/D(N) \mid N$, then we have $$I\bigg(\frac{N}{D(N)}\bigg) \leq I(N) < \frac{2N+D(N)}{N+D(N)}=\frac{2\bigg(\frac{N}{D(N)}\bigg)+1}{\bigg(\frac{N}{D(N)}\bigg)+1}.$$

Here is my question:

If $N$ is deficient-perfect, under what conditions does $$\frac{2\bigg(\frac{N}{D(N)}\bigg)}{\bigg(\frac{N}{D(N)}\bigg)+1} < I\bigg(\frac{N}{D(N)}\bigg)$$ hold?

It appears that it only holds when $D(N)=1$, i.e. when $N$ is almost perfect. Are there other conditions under which the inequality in the question holds?

Added April 26 2019 (7:25 PM Manila time)

The inequality $$\frac{2\bigg(\frac{N}{D(N)}\bigg)}{\bigg(\frac{N}{D(N)}\bigg)+1} < I\bigg(\frac{N}{D(N)}\bigg) < \frac{2\bigg(\frac{N}{D(N)}\bigg)+1}{\bigg(\frac{N}{D(N)}\bigg)+1}$$ holds if and only if $N/D(N)$ is almost perfect.

So here is my reformulated question:

If $N$ is deficient-perfect, when is $N/D(N)$ almost perfect?

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This is not really an answer, just some comments that would be too long to fit in the appropriate section.

The OEIS page on deficient-perfect numbers asserts that

If $2^{k+1} + 2^t - 1$ is an odd prime and $t \leq k$, then $n = 2^k(2^{k+1} + 2^t - 1)$ is deficient-perfect with $2n - \sigma(n) = 2^t$. In fact, these are the only terms with two distinct prime factors. (Tang et al.)

Suppose that $n = 2^k(2^{k+1} + 2^t - 1)$ with $2^{k+1} + 2^t - 1$ an odd prime and $t \leq k$. Then $D(n) = 2^t$, so that $$\frac{n}{D(n)} = 2^{k-t}(2^{k+1} + 2^t - 1).$$

Now we compute $$I\bigg(\frac{n}{D(n)}\bigg) = \frac{\sigma\bigg(2^{k-t}(2^{k+1} + 2^t - 1)\bigg)}{2^{k-t}(2^{k+1} + 2^t - 1)} = \frac{2^{k+1}+2^t}{2^{k-t}}\cdot\frac{2^{k-t+1}-1}{2^{k+1} + 2^t - 1}.$$

We need to check whether this is greater than $$\frac{2\bigg(\frac{n}{D(n)}\bigg)}{\bigg(\frac{n}{D(n)}\bigg)+1}=\frac{2^{k-t+1}(2^{k+1} + 2^t - 1)}{2^{k-t}(2^{k+1} + 2^t - 1) + 1}.$$

Assume that it is. Then we have $$\frac{2^{k+1}+2^t}{2^{k-t}}\cdot\frac{2^{k-t+1}-1}{2^{k+1} + 2^t - 1} > \frac{2^{k-t+1}(2^{k+1} + 2^t - 1)}{2^{k-t}(2^{k+1} + 2^t - 1) + 1}$$ which implies that $$\bigg(2^{k+1}+2^t\bigg)\bigg(2^{k-t+1}-1\bigg)\bigg(2^{k-t}(2^{k+1} + 2^t - 1) + 1\bigg) > \bigg(2^{k-t}\bigg)\bigg(2^{k+1} + 2^t - 1\bigg)\bigg(2^{k-t+1}(2^{k+1} + 2^t - 1)\bigg).$$ The LHS of the last inequality simplifies to $$-2^{3k-2t+2}+2^{4k-2t+3}+2^{2k-t+2}+2^{3k-t+2}-2^{k+t}+2^k-2^{2k+1}-2^t$$ while the RHS simplifies to $$2^{2k-2t+1}-2^{3k-2t+3}+2^{4k-2t+3}-2^{2k-t+2}+2^{3k-t+3}+2^{2k+1}.$$

We now test whether the LHS $-$ RHS $>0$.

We get $$\bigg(-2^{3k-2t+2}+2^{4k-2t+3}+2^{2k-t+2}+2^{3k-t+2}-2^{k+t}+2^k-2^{2k+1}-2^t\bigg)-\bigg(2^{2k-2t+1}-2^{3k-2t+3}+2^{4k-2t+3}-2^{2k-t+2}+2^{3k-t+3}+2^{2k+1}\bigg) = -2^{-2t}\bigg(-2^{2k+t+3}+2^{3k+t+2}-2^{k+2t}+2^{2k+2t+2}+2^{k+3t}+2^{2k+1}-2^{3k+2}+2^{3t}\bigg)$$ which is evidently negative since the second factor is definitely positive for $t \leq k$.

We therefore conclude that $n = 2^k(2^{k+1} + 2^t - 1)$ is not almost perfect for $2^{k+1} + 2^t - 1$ an odd prime and $t \leq k$. This finding agrees with Tang et. al.'s result that $D(n)=2^t \geq 2$ and the fact that even almost perfect numbers other than the powers of two, which takes the form $2^r b^2$ for $b$ an odd composite, must have at least three distinct prime factors.

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