The prime game & non-prime columns. 
The prime game:
We define $\Delta n$ to be the number of columns in the following table: $$
\overbrace{\begin{matrix}
1 & \color{green}2 & \color{green}3 & 4 & \color{green}5 \\
6 & \color{green}7 & 8 & 9 & 10 \\
\color{green}{11} & 12 & \color{green}{13} & 14 &  15  \\
16 & \color{green}{17} & 18 & \color{green}{19} & 20 \\
21 & 22 & \color{green}{23} & 24 & \ldots
\end{matrix}}^{\Delta n\text{ columns}} \text{$\qquad\Delta n =5$ in this case.}
$$
  And we also define a quantity $m$ that represents the number of columns skipped in the beginning: $$
\overbrace{\begin{matrix}
\leftarrow\overset{m}{}&{\rightarrow} & 1 & 2 & 3 \\
4 & 5 & 6 & 7 & 8 \\
9 & 10 & 11 & 12 &  13  \\
14 & 15 & 16 & 17 & 18 \\
19 & 20 & 21 & 22 & \ldots
\end{matrix}}^{\Delta n\text{ columns}}  \text{$\qquad m =2$ in this case.}
$$

Now, the questions: 


*

*Does there exist a non-trivial (i.e. other than $1,0$ & $1,1$) value for $\Delta n, m$ for which prime numbers all appear consecutively in the same column/columns and nowhere else? A situation like this: $${\begin{matrix}
x& p& x & p & x \\
x & p & x & p & x \\
x & p & x & p &  x  \\
x & p & x & p & x \\
x & p & x & p & \ldots
\end{matrix}}\qquad\begin{cases}x: & \text{nonprime.} \\ p: & \text{prime.} \\\end{cases}$$ $\rm\bf Op:$ I think the answer will be a "no" but we should first prove it, the thing that I'm struggling with for the moment.

*For which $\Delta n,m$ there exist a column (or more) who doesn't contain any prime? (other than the trivial one where we get a sequence of even numbers) 
 $\rm \bf Op:$ I think the answer will be a big number but I'm not sure. I need more tools to find a suitable proof. 

*What are all the pairs $\Delta n,m$ for which there exist a row that doesn't contain any prime?


I hope I'll find out the answer to these questions with your help.
 A: First question: this will never happen.  Each of your columns is an arithmetic sequence of the form
$$a,\,a+d,\,a+2d,\,a+3d,\,\ldots,\,a+kd,\,\ldots$$
where $d$ is what you have called $\Delta n$.  A sequence like this cannot contain only primes, it must have a composite number sooner or later.  The easiest way to (almost) see this is to consider when $k=a$: then the number is $a+ad$ which has a factor of $a$ and is therefore not prime.  Not quite right, however, as $a$ might be $1$ and then $1+d$ could possibly be prime.  So we have to be a bit more clever.  If we take $k=a(d+2)$ then the number is
$$a+kd=a+a(d+2)d=a(d+1)^2$$
which is definitely composite.
Second question: there are many cases where there will be a prime-free column.  Just make sure that $a$ and $d$ have a common factor bigger than $1$.  For example, in your notation take $m=0$ and $\Delta n=4$.  Then the fourth column will be $4,8,12,16,\ldots\,$: every number is a multiple of $4$ and therefore is not prime.
Third question: whatever numbers you pick, sooner or later there will be a prime-free row.  This is because you can find "gaps" in the primes as large as you like.  For example, take $\Delta n=5$.  The numbers
$$11!+2,\,11!+3,\,11!+4,\,\ldots,\,11!+11$$
are ten composite numbers in a row, see if you can explain why.  So there will be a row of $5$ in your table containing only composite numbers.
A: First note that each column will be an arithmetic progression with common difference $\Delta n$.
Dirichlet's theorem on primes in arithmetic progressions states; 

Let $n \in \mathbb{N}$, $n > 1$, and let $a \in \mathbb{N}$ be coprime to $n$. Then there are infinitely many primes $\equiv a$ (mod $n$).
  That is, there are infinitely many primes in the arithmetic progression $a, a+n, a+2n,...$

This is a high powered theorem (and the proof is "hard").
Using this, then, to have a column that contains no prime you would need $\Delta n$ and the first number in that column to have a non-trivial common factor - and note that if they do, then this factor will divide everything in the column, hence there will be no primes (except possibly for the first element of the column if it is itself prime, and is said common factor).
You should be able to use this to find examples of the $m, \Delta n$ required.
To have a column consisting entirely of primes, you'd need an infinitely long series of primes in arithmetic progression. The primes do contain arbitrarily long arithmetic progressions (see the Green-Tao theorm), but they are all finite - there have been many results in number theory on the distribution of the primes, I forget the proof or name of this one, but it should be findable if you look up the area.
A row containing no primes requires a list of $\Delta n$ consecutive composite numbers, starting at the correct point - to guarantee there is one starting to line up with a row, a list of $2 \Delta n$ consecutive composite numbers will suffice. It is possibly to prove that arbitrarly long strings of consecutive composite numbers exist (hint: try to find a string of length $k$ - consider factorials).
A: Note that $\Delta m$ only rotates and shifts the columns, so it does not really add something new. 
The answer to your first and third question is no, since there are arbitrarily large intervals containing no prime number at all (for a gap of size $n+1$ consider $n!, \dots, n!+n$).
Regarding your second question, let wlog $\Delta m = 0$ and consider the $k$-th column. Then:


*

*If $n$ and $k$ are not coprime,  there is exactly one prime in the $k$-th column if $k$ is prime and no prime otherwise, because any number in the column will be divisible by $(n,k)$.

*If $n$ and $k$ are coprime, there are infinitely many primes in the $k$-th column by application of Dirichlet's Theorem  (in fact the statement is equivalent).
We conclude that there is a prime in every column iff $n$ is a prime. The reason is that if $n$ is not a prime, there won't be a prime in the $2p$-th row for any prime $p \mid n$.
A: Every prime number has a table formulated by PN+(PNx6). This does generate columns of possible prime numbers (6n+or-1), plus two columns of composite numbers within the set of possible prime numbers, which work as a sieve when applied to the sequence of possible prime numbers. 
Here's a quick example figuring the prime numbers to 100 using these tables.
Possible prime numbers greater than 3:
. 5 . 7 .11 13 17 .19 23 25 29 31 
35 37 41 43 47 49 53 55 59 61
65 67 71 73 77 79 83 85 89 91
95 97
This is the Prime Number Table for 5. It identifies these composite numbers on the list of possible prime numbers: 25 35 55 65 85 95 
The PN Table for 7 identifies: 49 91
The PN Table for 11 identifies: 77
Using this sieve, there are just 9 calculations to eliminate the composite numbers up to 100, leaving only the prime numbers. For the Sieve of Eratosthenes, 113 calculations are needed.
