Checking for prime constellation

$n,n+2,n+4$ can't be all primes more than once because whatever is $n \bmod 3$ one of $n,n+2,n+4$ will be $\equiv 0 \bmod 3$.

Given $0<b_1< \ldots < b_k$ there is no obstruction for $n,n+b_1,\ldots,n+b_k$ being all primes (more than once) iff for every prime $p$ there is some $a_p \in 1\ldots p-1$ such that $a_p,a_p+b_1,\ldots,a_p+b_k$ are all coprime with $p$.

If $p > b_k+1$ there is the solution $a_p=1$.

Thus it suffices to look at the primes $p\le b_k+1$ which means there is no obstruction exactly when for some $A$ all of $A,A+b_1,\ldots,A+b_k$ are coprime with $(b_k+1)!$

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The natural generalization of the twin prime conjecture is that when there is no obstruction $n,n+b_1,\ldots,n+b_k$ are all primes infinitely many times.

The conjectured asymptotic (from the random model for the primes) for the number of such $n$ is $$\pi_b(x) \sim \sum_{n \le x} \prod_{p \le n^r} \frac{\sum_{a =1,\ p\, \nmid\, a (a+b_1)\ldots (a+b_k)}^{p-1} 1}{p} \sim C(b) \frac{x}{\ln^{k+1} x}$$ where for $p > b_k+1, \frac{\sum_{a =1,\ p\, \nmid\, a (a+b_1)\ldots (a+b_k)}^{p-1} 1}{p} = 1-\frac{b_k+1}{p}$

and $r =e^{-\gamma} \approx 0.56$ is the constant such that $\prod_{p \le n^r} \frac{p-1}{p} \sim \frac{1}{\ln n}$ (making the predicted $C(b)$ constants compatible with the PNT)

The random model for the primes is merely saying that when choosing $n$ randomly in $[1,x]$, for $p \le n^r$, then $n \bmod p$ can be considered uniformly distributed and more important the events $n \bmod p$ and $n \bmod q$ can be considered independent. Thus, under this model, the probability that $n$ is prime can be considered to be $\prod_{p \le n^r} \frac{p-1}{p}$ which by Mertens's theorem is $\sim \frac1{\ln n}$.