If $N = q^k n^2$ is an odd perfect number with $k > 1$ and $N = \frac{q(q+1)}{2} \cdot d$ for some $d > 1$, then what is $d$? By this answer, we know that every odd perfect number $N = q^k n^2$ can be written in the form
$$N = \dfrac{q(q+1)}{2} \cdot d$$
where $d > 1$.  (That is, an odd perfect number $N = q^k n^2$ is a nontrivial multiple of the triangular number
$$T(q) = \dfrac{q(q+1)}{2},$$
where $q$ is the Euler prime of $N$.)
If $k=1$, then it is easy to show that
$$d = D(n^2)$$
where $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of the non-Euler part $n^2$.
Here is my question:

If $N = q^k n^2$ is an odd perfect number with $k > 1$ and $N = \frac{q(q+1)}{2} \cdot d'$ for some $d' > 1$, then what is $d'$?

 A: As hinted by user25406 in a comment, setting up the quadratic in $q$, we obtain
$${d'}q^2 + {d'}q - 2N = 0.$$
In order for this equation to have (positive) integer solutions for $q$, we need the discriminant
$$D_1 := {d'}^2 + 8{d'}N$$
to be a perfect square.
This gives that
$$D_1 = {{d'}^2}\cdot(1 + 4q(q+1)) = \left({d'}(2q+1)\right)^2$$
is a perfect square, which is trivial.
Added January 25 2017
To clarify another comment, rewrite the equation
$${d'}q^2 + {d'}q - 2N = 0$$
as
$$q^2 + q - \dfrac{2N}{d'} = 0.$$
The discriminant is given by
$$D_2 := 1 + \dfrac{8N}{d'}.$$
user25406's (unproved) claim is that this is a perfect square only when $d' = N$.
A: This is more to the spirit of my expected answer for the original question, rather than the one presented in my other answer.
So we have an odd perfect number $N = q^k n^2 = {q(q+1)/2}\cdot{d'}$ (with $k>1$), where $d' > 1$ by Slowak's 1999 result.
Thus, $\frac{2N}{q(q+1)} = \frac{\sigma(N)}{q(q+1)} = d'$, so that
$$d' = \dfrac{\sigma(n^2)\sigma(q^k)}{q(q+1)} = {q^{k-1}}\cdot{\dfrac{\sigma(n^2)}{q^k}}\cdot{\dfrac{\sigma(q^k)}{q+1}}.$$
Now, I have shown elsewhere that
$$\dfrac{\sigma(n^2)}{q^k}=\frac{D(n^2)}{\sigma(q^{k-1})},$$
so that we obtain
$$d' = {q^{k-1}}\cdot{\dfrac{D(n^2)}{\sigma(q^{k-1})}}\cdot{\dfrac{\sigma(q^k)}{q+1}}.$$
Finally, we have
$$d' = \bigg(\dfrac{\sigma(q^k)}{\sigma(q)}\bigg)\cdot\bigg(\dfrac{D(n^2)}{I(q^{k-1})}\bigg)$$
or equivalently,
$$d' = {q^{k-1}}\cdot\bigg(\dfrac{I(q^k)}{I(q)I(q^{k-1})}\bigg)\cdot{D(n^2)},$$
where $I(x)=\sigma(x)/x$ is the abundancy index of the (positive) integer $x$.
Note that these last two formulas reduce to the case $d = D(n^2)$ when $k=1$, as expected.
