There is definitely something to be said, again, in terms of the compound arithmetic progression for James Maynard's concept of Large Gaps. Not only does this require us to go far beyond the simple Twin Prime sieve in the $\sigma$-ring of compound arithmetic progressions, it requires a description of De Polignac's Conjecture (1849) as a sequence in that ring to go beyond primorial descriptions of the largest prime gap under a magnitude or interval of consecutive composite numbers.
Conjecture (De Polignac, 1849). If $\mathbb{P}^{\gamma} = \{p_i, p_{i+1}\} \subset \mathbb{P}$ and $p_{i+1} -p_i= 2n$, for any given $n \in \mathbb{Z}^+$, there exist infinitely many $\mathbb{P}^{\gamma}$ satisfying the relation.
Proof is not part of the problem. Look up on Vixra if you want a more precise topological definition; a 2015 paper by a professor from Morocco is superbly concise and draws upon Fürstenberg's topological proof of the infinitude of primes in no uncertain terms. The numerical landmarks are obtained from compound arithmetic progressions, however, which I have already described in my answer to Prime Gaps in Residue Classes.
Conjecture. Let $\Delta \mathbb{P}_2$ be defined as the set of numbers for which $$\lambda \in \Delta \mathbb{P}_2 \implies \{6\lambda -1, 6\lambda +1\}\subset \mathbb{P} $$
Then if we let $T_C(r, m)$ be the composite topology in $[r]_m$
$$\Delta \mathbb{P}_2 = \mathbb{Z}^+ \setminus \bigcup_{r\in (\mathbb{Z}/6\mathbb{Z})^*}\{ T_C(r,6) \}$$
And expanding out to each matrix representation of the compound arithmetic progressions produced we can write $\Delta \mathbb{P}_2$ such that it is an element of the $\sigma$-ring of compound arithmetic progressions using the following shorthand (again, see the definition of the matrix representation):
$$\Delta \mathbb{P}_2 = \mathbb{Z}^+ \setminus \bigcup \{
M^{-1} \begin{pmatrix} -1 & n \\ 6 & 1 \end{pmatrix}, M^{-1} \begin{pmatrix} 1 & n \\ 6 & -1 \end{pmatrix}, M^{-1} \begin{pmatrix} 1 & n \\ 6 & 1 \end{pmatrix}, M^{-1} \begin{pmatrix} -1 & n \\ 6 & -1 \end{pmatrix} \}$$
For a gap size of 4, $\lambda \in \Delta \mathbb{P}_4$ implies that $\{6\lambda + 1, 6\lambda+5\} \subset \mathbb{P}$. The reasoning is that the use of negative numbers for the residue should be minimized and that this $6\lambda + 5 = 6(\lambda + 1) - 1$, so that the only difference between $\Delta \mathbb{P}_2$ and $\Delta \mathbb{P}_4$ is that one of the subtracted composite topologies is the translate in the representation.
$$\Delta \mathbb{P}_4 = \mathbb{Z}^+ \setminus \bigcup \{ T_C(1,6), T_C(-1,6) \oplus 1 \}$$
And $\lambda \in \Delta \mathbb{P}_6$ implies that $\{6\lambda - 1, 6\lambda +5\} \subset \mathbb{P}$ and ${6\lambda + 1} \not\in \mathbb{P}$ or $\{6\lambda + 1, 6\lambda + 7\} \subset \mathbb{P}$ and ${6\lambda + 5} \not\in \mathbb{P}$. So there is, in fact, two possible k-tuples, both with a composite region.
In terms of this structure, the composite topologies representing the composite region in the k-tuple ensure that the frontier prime elements are consecutive in the sequence of prime numbers, and therefore form an intersection of similarly translated composite topologies.
Therefore the results for $\Delta \mathbb{P}_6$ and the De Polignac sequence $\Delta \mathbb{P}_{2n}$ are as follows (in terms of composite topologies):
If $n\in \{1\pmod{3}\}, k := \frac{n-1}{3}$
$$\Delta \mathbb{P}_{2n} = \bigcap^{k}_{m=0}\{T_C(1,6) \oplus m \cap T_C(-1,6) \oplus (m+1)\} \setminus \bigcup\{ T_C(-1,6) \cup T_C(1,6)\oplus k\}$$
If $n\in \{2\pmod{3}\}, k := \frac{n-2}{3}$
$$\Delta \mathbb{P}_{2n} = \bigcap^{k}_{m=0}\{T_C(1,6) \oplus (m+1) \cap T_C(-1,6) \oplus (m+1)\} \setminus \bigcup\{ T_C(1,6) \cup T_C(-1,6)\oplus (k+1)\}$$
And finally if $n \in \{ 0\pmod 3\}, n>0$ there are again, two ways to form the k-tuple for the gap, so in terms of composite toplogies:
$$\Delta \mathbb{P}_{2n} = \{\bigcap^{k}_{m=0}\{T_C(1,6) \oplus m \cap T_C(-1,6) \oplus (m+1)\} \cap \{T_C(1,6)\oplus k\}\} \setminus \bigcup\{ T_C(-1,6) \cup T_C(-1,6)\oplus (k+1)\} \cup $$
$$ \{\bigcap^{k}_{m=0}\{T_C(1,6) \oplus (m+1) \cap T_C(-1,6) \oplus (m+1)\} \cap \{T_C(-1,6)\oplus (k+1)\}\} \setminus \bigcup\{ T_C(-1,6) \cup T_C(-1,6)\oplus (k+1)\} $$
Which is what I was able to derive for the general form of the De Polignac Sequence in the above mentioned ring. And it is possible to analyze the infima of each element of the sequence or for a gap size that you are more curious about, or if you want to find a sequence of consecutive composite numbers . That's how it's done. Large gaps is a difficult problem. The notation looks like the machine language of a computer and it could take a volume to try to decompile it. But $\phi(6) = 2$, so there are at most 2 CAPs per composite topology and then in the long hand $inf \bigcup{[ax+b]^+_{(cx+d)}} = (a+c)x+(b+d)$, it is possible to use the case where $x:=1$ so that the figure becomes $a+b+c+d$, where the matrix form is $\begin{pmatrix} -a & n-b \\ c & d \end{pmatrix}$
Have fun with that, for now.