Infinite prime cylinders? Define a prime $n$-circle as a circular sequence of
$n$ distinct natural numbers such that adjacent elements sum to a prime
(including $n^{\textrm{th}}$ $+$ $1^{\textrm{st}}$).
For example:
$$6, 1, 2, 5, 8, 9, 4, 7 \;.$$
This is a variation on the prime circles counted in
the integer sequence
A051252,
in that I am not insisting that the numbers be drawn from
$\{1,2,\ldots,n\}$. (The above circle misses $3$.)
Define a prime $n$-cylinder as an aligned stacking of prime
circles, such that adjacent numbers vertically also sum to a prime.
(However, no wrap-around requirement from top to bottom.)
Further, every circular rung of the cyclinder should be a
distinct permutation—no repeats allowed (also disallowing
reversals).
For example, here is a prime octagonal-cylinder consisting of 
four prime circles:

          


And here displayed as a matrix:
$$
\left(
\begin{array}{cccccccc}
 7 & 4 & 3 & 8 & 9 & 2 & 5 & 6
   \\
 16 & 15 & 8 & 9 & 4 & 3 & 2 &
   1 \\
 7 & 4 & 9 & 8 & 3 & 2 & 15 &
   16 \\
 6 & 1 & 2 & 5 & 8 & 9 & 4 & 7
   \\
\end{array}
\right)
$$

Q. Do there exist infinitely tall prime $n$-cylinders,
  for each $n$?

The answer should
be the same whether the cyclinder is
infinite just in one direction, or bi-infinite, extending infinitely
in both directions.
Although it seems relatively easy to add rungs in a greedy fashion,
I don't see how to prove that this approach can extend infinitely.
For example, here is how one might get "stuck."
Suppose you are adding the last number $x$ to the
top prime circle. The three numbers adjacent to $x$ (left, right, below)
could be $1$, $3$, and $5$. But the only number $x$ such
that each of $\{x+1, x+3, x+5 \}$ is prime is $x=2$, which may already
have been used in that top circle.

Let's denote the numbers in rung $j$ by $a(i,j)$ where $i$ runs from $1$ to $n$. So the conditions for a infinite (in one direction) cylinder are:


*

*A. $a(1,j) + a(n,j)$ is prime for $j \ge 1$

*B. $a(i,j) + a(i-1,j)$ is prime for $i = 2 \dots n$, for $j \ge 1$

*C. $a(i,j) + a(i,j-1)$ is prime for $i=1 \dots n$, for $j \ge 2$
Clearly $n$ must be even and the numbers in each rung must alternate between even and odd.
If we have constructed up to rung $k$ then we can construct most of rung $k+1$ as follows:
$a(i,k+1) = a(i+1,k)$ for $i=1 \dots n-1$
This meets conditions B and C on rung $k+1$, up to $i=n-1$. To meet the remaining conditions we just need to choose $a(n,k+1)$ so that:


*

*$a(n,k+1) + a(n-1,k+1)$ is prime

*$a(n,k+1) + a(n,k)$ is prime

*$a(n,k+1) +a(1,k+1)$ is prime


and also $a(n,k+1) \gt \max(a(i,k))$ for $i=1..n$ to prevent a repeat. And since $a(n-1,k+1)=a(n,k)$, the three constraints above in fact reduce to two.
Since we can make $a(n,k+1)$ as large as we like, we simply find two primes $p$ and $q$ such that $p,q \gt 2\max(a(i,k))$ and $p-q=a(n-1,k+1)-a(1,k+1)$, and then set
$a(n,k+1) = p - a(n-1,k+1) = q-a(1,k+1)$
As an example, an infinite cylinder with initial rung $\left(
\begin{array}{cccccccc}
 6 & 1 & 2 & 5 & 8 & 9 & 4 & 7
   \\
\end{array}
\right)$ could be
$\left(
\begin{array}{cccccccc}
 6 & 1 & 2 & 5 & 8 & 9 & 4 & 7
   \\
 1 & 2 & 5 & 8 & 9 & 4 & 7 & 12
   \\
 2 & 5 & 8 & 9 & 4 & 7 & 12 & 17
   \\
 5 & 8 & 9 & 4 & 7 & 12 & 17 & 24
   \\
 8 & 9 & 4 & 7 & 12 & 17 & 24 & 29
   \\
 . & . & . & . & . & . & . & .
   \\
\end{array}
\right)$
... and if you want a doubly infinite cylinder you can do the same trick in the other direction ...
$\left(
\begin{array}{cccccccc}
. & . & . & . & . & . & . & .
   \\
33 & 28 & 25 & 6 & 1 & 2 & 5 & 8
   \\
28 & 25 & 6 & 1 & 2 & 5 & 8 & 9
   \\
25 & 6 & 1 & 2 & 5 & 8 & 9 & 4 
   \\
 6 & 1 & 2 & 5 & 8 & 9 & 4 & 7
   \\
 1 & 2 & 5 & 8 & 9 & 4 & 7 & 12
   \\
 2 & 5 & 8 & 9 & 4 & 7 & 12 & 17
   \\
 5 & 8 & 9 & 4 & 7 & 12 & 17 & 24
   \\
 8 & 9 & 4 & 7 & 12 & 17 & 24 & 29
   \\
 . & . & . & . & . & . & . & .
   \\
\end{array}
\right)$
