Do the Apollonian and Markoff-Hurwtiz problems share the same "jumping" movement? That would depend on one's interpretation of "share." There is a link; one that my research followed.
I was first introduced to the Hurwitz equations via the Markoff equation; and the fractal nature of the exponent of growth lead me to the work of Boyd on the Apollonian Packing Problem. I later heard about a certain class of K3 surfaces ... those surfaces that are smooth $(2,2,2)$ forms in $P\times P\times P$. (Note that $P\times P\times P$ is not the same as $P^3$.) These are surfaces in three projective variables whose defining equation is quadratic in each variable. They therefore have the same "jumping" action as the Markoff equation. These surfaces have the advantage of being smooth, so the jumps end up being automorphisms of the surface. The automorphisms act on the group of divisors on the surface (the Picard group), and this action shares a lot of similarities with the Apollonian packing.
The Apollonian packing can be thought of this way: The bilinear form that you mention in your note is a Lorentz product. That is, we can write it as $X^tJX=0$, where $X=(w,x,y,z)$ and $J$ is a symmetric matrix with signature $(3,1)$. (That is, $J$ has $3$ positive eigenvalues and one negative eigenvalue.) Thus, a surface of the form $X^tJX=-1$ is a hyperboloid of two sheets, and one sheet is a model of 3D hyperbolic geometry, with distance defined by $\cosh(|XY|)=-X^tJY$. Three dimensional hyperbolic space has the Poincare upper half space model, and planes in this model are (Euclidean) hemi-spheres perpendicular to the boundary. A plane is therefore represented by a circle. The "jumps" used in the Apollonian packing is a subgroup of the group of all such jumps, and is a subgroup of isometries in the hyperbolic space. The packing is an orbit of one plane under the action of this group. Because the fundamental domain for the group of "jumps" has infinite volume, the packing has an associated fractal.
For the K3 surfaces, the Lorentz product is the intersection pairing, which has signature $(1,3)$ (if the Picard group has dimension $4$). The group of automorphisms on the K3 surface (analogs of the jumps for the Markoff equation) acts as isometries (analogs of jumps for the Apollonian packing) on the underlying hyperbolic space. The orbit of a circle for these cases are not always tangent, as they are in the Apollonian case, but are sometimes perpendicular. An example is pictured on my homepage (http://faculty.unlv.edu/baragar/). Note that one of the jumps in the Picard space came from an automorphism that isn't exactly a "jump" on the surface, though it has a very interesting interpretation.
There is no K3 surface, though, that is associated to the Apollonian packing.
Proposition 1: The number of integer solutions of the equation
$$
\sum_{i=1}^{k}x_i = N
$$
where $x_i\geq n_i$ for $i=1,\ldots,k$, is given by
$$
{\small
\binom{N+k-1-n_1-n_2-...-n_k}{k-1}
}
$$
if the upper index is non-negative and zero otherwise.
In the formula above, $\binom{.}{.}$ stands for the generalized binomial coefficients.
Now, to tackle the problem as stated, you need to apply Proposition 1 and invoke the inclusion-exclusion principle, in the following sense:
For $i=1,...,k$, set as
$q_i$: the property of a solution of Proposition 1, to satisfy the condition
$$
x_i> m_i \Leftrightarrow x_i\geq m_i+1
$$
If we denote:
- $N(q_i)$, the number of solutions (provided by Prop. 1) satisfying property $q_i$,
- $N(q_i q_j)$, the number of solutions (provided by Prop. 1) satisfying both properties $q_i$, $q_j$,
... and generally:
- $N(q_{i_1}q_{i_2}... q_{i_s})$, the number of solutions (provided by Prop. 1) satisfying all properties $q_{i_1}$, $q_{i_2}$, ..., $q_{i_s}$,
then we get -applying Prop. 1- that:
$$
{\small
N(q_1)=\binom{N+(k-1)-1-m_1-n_2-...-n_k}{k-1} \ \ \ or \ \ \ N(q_1)=0
}
$$
$$
{\small
N(q_2 q_3)=\binom{N+(k-2)-1-n_1-m_2-m_3-n_4-...-n_k}{k-1}
\ \ \ or \ \ \ N(q_2 q_3)=0}
$$
... and generally:
$$
{\small
N(q_{i_1}q_{i_2}... q_{i_s})=\binom{N+(k-s)-1-\sum_{i\notin I} n_i-\sum_{i\in I} m_i}{k-1} \ \ \ or \ \ \ N(q_{i_1}q_{i_2}... q_{i_s})=0
}
$$
where $s$ is the number of properties, $I=\{i_1, i_2, ..., i_s\}\subseteq \{1,2,...,k\}$ and the $N(..)$ function takes zero values whenever the upper index becomes negative.
Now all you need to do to obtain a compact formula for the number of solutions satisfying your constraints, is to apply the inclusion-exclusion principle to determine the number of solutions produced by Proposition 1, which have none of the properties $q_i$ for $i=1,2,...,k$.
This is given by
$$
\binom{N+k-1-\sum n_i}{k-1}-\sum_{i=1}^{k} N(q_i)+\sum_{k\ \geq j > i\geq 1} N(q_i q_j)-...+ \\ +(-1)^s\sum_{k\ \geq i_s>...>i_1\geq 1} N(q_{i_1}q_{i_2}... q_{i_s})+.... +(-1)^k N(q_{1}q_{2}... q_{k})
$$
where in the above formula $s$ is the number of properties and
$$
\sum_{k\ \geq j > i \geq 1}=\sum_{i=1}^{k-1}\sum_{j=i+1}^{k}
$$
... etc.
Example: As an example of application of the previous method, consider the following special case of the OP:
Find the number of (positive) integer solutions of the equation
$$\sum_{i=1}^{k}x_i = N$$ for some positive integer $N$, given the constraints $1\leq x_i\leq \alpha$ for $i=1,\ldots,k$
The method described above gives:
$$
{\small
\binom{N-1}{k-1}-\binom{k}{1}\binom{N-\alpha-1}{k-1}+\binom{k}{2}\binom{N-2\alpha-1}{k-1}-\binom{k}{3}\binom{N-3\alpha-1}{k-1}+\binom{k}{4}\binom{N-4\alpha-1}{k-1}-\cdots
}
$$
where $\binom{..}{..}$ stands for the generalized binomial coefficients and the summation halts when zero terms appear.
Best Answer
As far as I know, the only significant result to speed up these calculations is that $E_2(\frac{p}{q}) = \frac{1}{2}|\lbrace d: d | q^2, d \equiv -q (mod p) \rbrace|$, where $E_2(p/q)$ represents the number of decompositions into 2 unit fractions, and each matching $d$ represents the decomposition $\frac{p}{q} = \frac{qp}{q(q+d)} + \frac{dp}{q(q+d)}$. (Take floor() or ceil() depending on whether you want to allow repeats.)
When I've coded this in the past, I called one of 4 different functions depending on a) whether $p=1$ or not, and b) whether $q/p \ge min$ or not, where $min$ is the greatest denominator I'm already using. When $p=1$ and $q \ge min$, in particular, we can just calculate $\tau(q^2)/2$ from the factorisation of $q$; in the other cases I actually walked the factors from $q/p$ to $\sqrt{q}$.
So: yes, you can count the number of matching sets without generating the 7 elements of each set, but computationally the elements are just a whisker away.