Which constellations of primes recur forever? Having derived much joy and learning from the answers I have received to four previous questions, let me ask one more. 
Let a constellation of primes be a set of primes that stand in certain fixed distances to one another. Thus {3,5}, {11,13}, {41,43} all exemplify the same constellation {n,n+2}. And {11,13,17,19,23} and {101,103,107,109,113} both exemplify the same constellation {n,n+2,n+6,n+8,n+12}. We may wonder which such constellations of primes recur forever. 
My understanding is that no constellation has yet been shown to be recurring forever. It is believed that twins of the form {n,n+2} recur forever, but this has not yet been demonstrated.
My question concerns the identification of constellations of primes as non-recurring. A simple example is the constellation {n,n+2,n+4} which we know is non-recurring because, for n>3, one of the three numbers must be non-prime by virtue of being divisible by 3.
My question: is this basic method the only one we have for demonstrating that some constellation of primes does not recur beyond a certain point? Or have other methods of proof been used and, if so, what are they?
If I understand the current state of knowledge correctly, we do not know of any constellation of primes -- not including {n} and trivial cases like {n,n+1} or {n,n+2,n+4} -- whether it does or does not recur forever. A lot of effort is currently being put into showing that certain constellations -- e.g., {n,n+2} -- do recur forever. We might also put effort into finding ways of showing that certain constellations do not recur forever. Has it ever been shown, of any constellation of primes that recurs at least (say) 20 times that it does not recur forever?
 A: The answer to your first question is definitely "yes".  There has never been a prime constellation proven to be occur finitely many times by any means other than a simple argument of the type "at least one number in the group must be divisible by $p$" for some prime $p$.  In fact, it is widely believed that any constellation that eludes this argument for all $p$ must occur infinitely many times, and furthermore that we can predict asymptotically how often it occurs.  This is the content of the prime $k$-tuples conjecture of Hardy and Littlewood: http://mathworld.wolfram.com/k-TupleConjecture.html.  The type of constellation you're asking for would be a counterexample to the prime $k$-tuples conjecture, and since that's still open, no such constellation has ever been proven to exist.
Your last question about a non-recurrent constellation which occurs a finite, but sizable number of times is interesting, though.  At first I was skeptical that such a constellation could occur even twice (let alone 20 times), but since I couldn't prove it impossible, I looked it up and it turns out that $\{n, n+2, n+8, n+14, n+26\}$ does occur twice ($n=3$ and $n=5$).  So $20$ times is probably possible, but the constellation would have to contain a large number of terms, which might place it beyond the reach of practical computation.
EDIT: Here's a construction that I think works to show that prime $k$-tuples conjecture $\implies$ for any $m \in \mathbb N$ there exist constellations that occur exactly $m$ times.
Lemma [Conditional on $k$-tuples]: Let $(a_1,a_2,a_3,\ldots, a_r)$ be an admissible set with $a_1 = 0$, and let $p$ be a prime greater than $r$.  Fix $b \in \mathbb Z$ so that $b \not\equiv -a_i \pmod p$ for all $i=1,\ldots,r$.  Then there are infinitely many prime constellations of the form $(q + a_1,  q+a_2, \ldots, q+a_r)$ having $q \equiv b \pmod p$.
Proof:  We want to augment the admissible set with additional terms that force $q \equiv b \pmod p$.  Let $P$ be a positive integer divisible by all primes strictly less than $p$, and let $Q$ be an inverse of $P$ modulo $p$.  Let $\mathcal C$ be the set of all congruence classes mod $p$ except for $-b$ and $a_1,\ldots, a_r$.  For any $c\in \mathcal C$, the integer $cPQ$ is divisible by all primes less than $p$, and congruent to $c$ modulo $p$.  It's not hard to see that $\{a_1,a_2,a_3,\ldots, a_r\} \cup \{cPQ : c\in \mathcal C\}$ is also admissible.  The choice of $\mathcal C$ forces all but finitely many prime constellations of this augmented form to have $q \equiv b \pmod p$.
Construction:
Start with a prime arithmetic progression of length $2m-1$, say $p, p+d, p+2d, \ldots, p + (2m-2)d$.  Without loss of generality we can assume that $(0,d,2d,\ldots, (2m-2)d)$ is admissible, and that $p$ is the first occurrence of that constellation.
Let $q$ be the middle prime of this progression, $p + (m-1)d$.  Now, take the subconstellation $(0,d,\ldots,(m-1)d)$ and use prime $k$-tuples to find $q-m$ additional prime constellations of this type (more details on this below).  In particular we can ensure that every element of the matrix below is prime:
$$\begin{pmatrix}
p & p+d & \cdots & p+(m-1)d \\
p+d & p+2d & \cdots & p+md \\
\vdots & \vdots& \ddots& \vdots \\
p+(m-1)d & p+md & \cdots & p+(2m-2)d \\
p_{m+1} & p_{m+1}+d & \cdots & p_{m+1}+(m-1)d \\
p_{m+2} & p_{m+2}+d & \cdots & p_{m+2}+(m-1)d \\
\vdots & \vdots& \ddots& \vdots \\
p_q & p_q+d & \cdots & p_q+(m-1)d 
\end{pmatrix}.$$
Note that the matrix is more uniform once we define $p_i := p+(i-1)d$ for $i=1\ldots, m$.  By the Lemma (noting that $q > m$), we can without loss of generality choose $p_{m+1},\ldots,p_q$ so that the first column of the matrix forms a complete residue system mod $q$.
Finally, construct the new constellation $\mathcal C = (p_1, p_2,\ldots, p_q)$, i.e. the first column of the matrix.  Clearly $\mathcal C$ occurs at least $m$ times, once for each column of the matrix.  However, since $C$ forms a complete residue system mod $q$, each occurrence of the constellation $\mathcal C$ must have one of its primes equal to $q$.  But since $p_i > q$ for each $i>m$, this can only happen at one of the first $m$ positions, and we've already accounted for those.  So $\mathcal C$ occurs exactly $m$ times.
