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.
 A: Since $\sigma(n)>0$ we have $d<2N$, together with $(2/3)N^2<d$ we get $(2/3)N^2<2N$ which is verified only for $0< N< 3$, which leaves only 2 cases to be checked, namely $1$ and $2$.
edit:After the question being asked has been changed the first part of my answer is irrelevant. I can't completely answer the new question, but I can show some solutions.
Consider a pair of twin primes $p$ and $p+2$ and their product $N=p^2+2p$.$\sigma(N)$ is clearly given by $1+p+p+2+p^2+2p$, that is $\sigma(N)=p^2+4p+2$.  From $\sigma(N)=2N-d$ we get $d=2(p^2+2p)-(p^2+4p+2)=p^2-3$
Substituting into the inequality $(2/3)N<d<N-\sqrt{N}$ we get
$$\frac{2}{3}(p^2+2p)<p^2-3<p^2+2p-\sqrt{p^2+2p}$$
The left inequality is satisfied for $p>2+\sqrt{13}$ while the right one is satisfied for $p\ge0$.
So as long as $p$ and $p+2$ are both prime and $p\ge11$ their product is a number with the required deficiency (whether there are infinite pairs of twin primes is an exercise left to the reader :P). 
For example consider $p=29$, that gives $N=899$, $\sigma(N)=960$,  $(2/3)N=599.33...$, $d=838$ and $N-\sqrt{N}=869.01...$
An analogous reasoning shows that if $p$ and $p+4$ are both primes and $p\ge7$ then their product is another solution, for example $77=7\times11$ and $221=13\times 17$. I suspect this can be generalized to more semiprimes, maybe all of them bigger than a certain number?
A: 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:
$$5\sqrt{N} + 5 < 9\sqrt{N} - 3 \implies 8 < 4\sqrt{N} \implies 2 < \sqrt{N} \implies 4 < N.$$
