<s>[INST] How Deficient a Number is? (Finding numbers having a certain deficiency) This question was edited, in particular equations were corrected:
A number N is said to be deficient by an integer $d$ if:
$\sigma(N)=2N-d$
Note that powers of 2 are deficient by 1.
While a prime $p$ is deficient by $p-1$
My question will be this:
Is there an integer $N$ that is deficient by an integer $d$ that lies between $(2/3)N$ and $N-\sqrt{N}$, that is
$(2/3)N<d<N-\sqrt{N}$ and can we characterize them?
Thanks a lot for your help.
 [/INST] Let us try using the following criterion:
Let $1 < d = 2N - \sigma(N)$.  Then
$$\frac{2N}{N + d} < I(N) < \frac{2N + d}{N + d}$$
where $I(N)=\sigma(N)/N$ is the abundancy index of $N$.
Since you require
$$\frac{2}{3}N < d < N - \sqrt{N},$$
then we have
$$\frac{2}{3} < \frac{d}{N} = 2 - I(N) < 1 - \sqrt{\frac{1}{N}}$$
so that we obtain
$$\frac{2}{2 - \sqrt{\frac{1}{N}}}< \frac{2}{1 + \frac{d}{N}} = \frac{2N}{N+d} < I(N) < \frac{2N+d}{N+d} = \frac{2+\frac{d}{N}}{1+\frac{d}{N}} < \frac{3 - \sqrt{\frac{1}{N}}}{\frac{5}{3}} = \frac{3}{5}\left(3 - \sqrt{\frac{1}{N}}\right).$$
However, from
$$\frac{2}{3} < \frac{d}{N} = 2 - I(N) < 1 - \sqrt{\frac{1}{N}}$$
you have the bounds
$$1 + \sqrt{\frac{1}{N}} < I(N) < \frac{4}{3},$$
so that
$$\frac{2}{2 - \sqrt{\frac{1}{N}}}<I(N)<\frac{4}{3}$$
and
$$1 + \sqrt{\frac{1}{N}}<I(N)<\frac{3}{5}\left(3 - \sqrt{\frac{1}{N}}\right).$$
The first inequality yields:
$$6 < 8 - 4\sqrt{\frac{1}{N}} \implies 4\sqrt{\frac{1}{N}} < 2 \implies 2 < \sqrt{N} \implies 4 < N.$$
The second inequality yields:
$$5\sqrt{N} + 5 < 9\sqrt{N} - 3 \implies 8 < 4\sqrt{N} \implies 2 < \sqrt{N} \implies 4 < N.$$
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