Given a quadruple of primes (distinct from $5,7,11,13$) note that:
The quadruplet $(11,13,17,19)$ I can write it as $(15n-2^2, 15n-2^1, 15n+2^1, 15n+2^2)$, $n=1$.
For the quadruplet $(101, 103, 107, 109)$, the same thing happens but this time for $n=7$.
My question is, why is it that the quadruple house can be written as $15n\pm 2^q$?
more generally, there are more than k tuples that can be written as $\alpha n\pm 2^q$?
Suppose $(p,p+2,p+6,p+8)$ is a prime quadruple. Since $p$ and $p+2$ are prime, $p+1$ and hence $p+4$ must be divisible by $3$.
If $p+1$ were divisible by $5$, also $p+6$ would be divisible by $5$ which is impossible since $p+6$ is assumed prime. Similarly, if $p+3$ were divisible by $5$, so would $p+8$ be. This is again impossible, so we have four consecutive numbers $p,p+1,p+2,p+3$ none of which are divisible by $5$. This is only possible if $p+4$ is divisible by $5$. As we have found it to be divisible by $3$ as well, it is divisible by $15$.
Note that this proof does not work if $p$ is $3$ or $5$. Note also the similarity with twin primes, which are always $(6n-1,6n+1)$ for some $n$ (with the obvious exception of $3$ and $5$).