This question concerns the generalization of certain characteristics of twin primes to a broader class of pairs of primes, and whether the generalized formulation might be used to provide insights into the twin prime conjecture.
Terminology: gaps vs spacings. The difference between two consecutive primes is canonically referred to as a gap: $p_{n+1}-p_n=g_n$. I refer to the difference between any two primes, whether or not consecutive, as a spacing: $p_a-p_b=s(p_a,p_b)$. For pairs of odd primes, gaps and spacings are always even. Since there are infinitely many primes of the form $6m-1$ (same for $6m+1$), many spacings will be multiples of $6$.
My broadest question is this: Is it known whether there are particular spacings between primes which occur infinitely often? My follow on question at the end will concern the possible infinitude of the particular spacings described next. My sense is that based on Dirichlet, which establishes that there are infinitely many primes in an arithmetic progression $a+kd$ with fixed $a,d$ and increasing $k$, there will be an abundance of spacings that are multiples of $d$, and that at least some multiples must occur infinitely many times. Has this been proved?
Background on twin primes: Other than the pair $3,5$, twin primes have the form $6m\pm 1$ where each of $6m-1,6m+1$ is a prime. The product of twin primes is a semiprime having the form $36m^2-1$. Not every $m\in \mathbb Z_+$ gives rise to twin primes, so not every number of the form $36m^2-1$ is a semiprime. Values of $m$ that give rise to twin primes are described in OEIS A002822.
Generalization to certain other pairs: Consider the set $\{s_n|s_{2k}=18k-1; \ s_{2k+1}=18k+1; \ k \in \mathbb N_0\}$. Note that every value of the index $n$ is associated with a specific value of $k_n=\lceil \frac{n-1}{2}\rceil$. The coefficient $18$ ensures that $(s_n^2-1)=324k_n^2\pm 36k_n \equiv 0 \bmod 36$. Let $t_n=\frac{s_n^2-1}{36}=9k_n^2\pm k_n$ with the appropriate sign, i.e. $-$ when $n$ is even and $+$ when $n$ is odd.
Consider pairs $6m\pm s_n \text{ where } (6m-s_n)>0$. Members of such pairs may or may not be prime, differ by a spacing of $2s_n$, and are of different parity (i.e. $\pm 1$) $\bmod 6$. Spacings of $2s_n$ are not multiples of $6$. The products of such pairs, $36m^2-s_n^2$, are $\equiv -1 \bmod 6$. I am interested in conditions sufficient for such products to be semiprimes.
One further restriction must be observed. In the case of twin primes, there are no other primes between $6m-1 \text{ and } 6m+1$. The same cannot be assured for pairs $6m-s_n \text{ and } 6m+s_n$, and in fact it is frequently the case that primes exist between the members of such pairs. Thus, if $36m^2-s_n^2$ is not a semiprime, $6m-s_n$ may nonetheless be its smallest prime factor, with $6m+s_n$ being composite. To ensure that we are dealing with cases in which the smallest prime factor of $36m^2-s_n^2$ is smaller than $6m-s_n$, we will limit ourselves to pairs for which $\sqrt{6m+s_n}<6m-s_n$, ensuring that if $6m+s_n$ is composite, it will have at least one prime factor smaller than $6m-s_n$. This restriction does not eliminate a large fraction of pairs from consideration, as the inequality is challenged only when $s_n$ is similar in magnitude to $6m$. For any given $s_n$, there are infinitely many acceptable values of $m$.
Characterization of prime pairs: If a number of the form $36m^2-s_n^2$ is not a semiprime, then it can be factored as $(6a\pm 1)(6b\mp 1)$ for some $a<m, a<b$. It is sufficient, although not necessary, to confine ourselves to $6a\pm 1$ being prime (proof omitted here). $$36m^2-s_n^2=(6a\pm 1)(6b\mp 1)=36ab\pm 6b \mp 6a -1 \\ 36m^2-(s_n^2-1)=36ab\pm 6b \mp 6a=36ab \pm 6(b-a) $$ Since LHS is divisible by $36$, it must be the case that $b-a=6c \Rightarrow b=a+6c$. Dividing through by $36$ and substituting, including $t_n$ for $\frac{s_n^2-1}{36}$, we get $$m^2=a^2+6ac\pm c + t_n \\ m^2-a^2=c(6a\pm 1) + t_n$$ The sign in $6a\pm 1$ in the last line tracks to the same number postulated in the opening. The implication is that if $36m^2-s_n^2$ is not a semiprime, then $m^2-a^2 \equiv t_n \bmod (6a \pm 1)$ for some $a$. Stated the other way, if $36m^2-s_n^2$ is a semiprime, then for all $a<m$, $m^2-a^2 \not \equiv t_n \bmod (6a \pm 1)$. In order to make clear what is transpiring behind all of the symbols, I include a table of the first several instances:
$$\begin{array}{|c|c|c|c|c|c|} \hline n & k_n & s_n & t_n &\exists \text{ Prime pairs} & \text{When } \forall a<m \\ \hline 1 & 0 & 1 & 0 & 6m \pm 1 & m^2-a^2 \not \equiv 0 \bmod (6a \pm 1) \\ \hline 2 & 1 & 17 & 2 & 6m \pm 17 & m^2-a^2 \not \equiv 2 \bmod (6a \pm 1) \\ \hline 3 & 1 & 19 & 4 & 6m \pm 19 & m^2-a^2 \not \equiv 4 \bmod (6a \pm 1) \\ \hline 4 & 2 & 35 & 10 & 6m \pm 35 & m^2-a^2 \not \equiv 10 \bmod (6a \pm 1) \\ \hline 5 & 2 & 37 & 14 & 6m \pm 37 & m^2-a^2 \not \equiv 14 \bmod (6a \pm 1) \\ \hline \end{array}$$
If there are infinitely many pairs of primes that have a spacing of $2s_n$, that means there is no $m_{\text{max}}$ such that for $m>m_{\text{max}}$, there is always some $a$ such that $m_2-a^2\equiv t_n \bmod (6a \pm 1)$. For $s_n=1$, this is equivalent to the twin prime conjecture.
My follow on question: For any $n$, can it be proved that prime pairs with spacings of the form $2s_n$ occur infinitely often?