Finite number of consecutive prime numbers

Question

Suppose we have $n$ consecutive prime numbers $p_k$ with $n \in\mathbb{N}$ such that $p_k-p_{k-1}=m$ and $k=2..n$. Is it possible to find $m$ and $n$ in order to have a finite number of these consecutive primes? For example, if we put $m=6,n=3$, for the first $23$ primes, we get only the triplet $(47,53,59)$ satisfying this condition. Obviously, for example, for $n=3,m=2$ (twin primes), there are only three consecutive primes $(3,5,7)$. Is this the only case we can have a finite number of consecutive primes with $n\gt 2$? Thanks.

Answer 1

A RATHER LONG COMMENT

For consecutive primes larger than $5$ in an arithmetic progression, the common increment must be a multiple of $6$, as those primes have a form $6t\pm 1$. Any five consecutive positive integers of the form $6t-1$ or $6t+1$ (or $6st \pm 1$ where $5 \not \mid s$) will have one member that is divisible by $5$, limiting sequences of primes to at most $4$ in length.

You can address this by making $5$ a factor of the increment, looking for strings among integers of the form $30t \pm r$ where $r \in \{1,7,11,13\}$, but then strings of $7$ such numbers will contain at least one member divisible by $7$. You could move to numbers of the form $210t \pm r$, but you will appreciate that as the increment of the arithmetic sequence gets larger, the likelihood of finding strings of consecutive primes diminishes.

This does not prove or even suggest that indefinitely long strings of consecutive prime numbers in an arithmetic sequence cannot be found, only that one very quickly is forced to look among very large numbers to have any chance of finding them.