Prime chains with large gaps

Question

It is well known that the gap between consecutive primes is unbounded. Is this still true for a chain of consecutive primes ?

More Formally : Is the following statement true for all natural numbers m and n ?

There are m consecutive primes $a_1,...,a_m$ , such that all the gaps are greater than n (this means $a_{k+1}-a_k>n$ for all k with 1 <= k <= m-1) ?

I also heard about primes in arithmetic progressions, but I always wondered if the primes must be consecutive in such progressions.

Can any of the known properties of the prime-numbers help to answer this question ?

Answer 1

Yes, this is known. It feels a bit like squashing a fly with a B-52, but there is a beautiful theorem of Shiu (Strings of Congruent Primes, 2000) which seems to cover everything you want. A direct consequence is that for any $n,m$ the following is true:

There are $m$ consecutive primes $a_1,\ldots,a_m$ that are all congruent to $1$ mod $n$.

So not only are the gaps between these primes at least $n$, but they all lie within a prescribed arithmetic progression. (It's important not to confuse this with $a_1,\ldots,a_m$ forming an arithmetic progression: a result of such strength is not yet known.)

EDIT: Another approach is the following: for any fixed $n$ let $P_n$ be the set of primes whose gap to the next prime is $\le n$. It is known from sieve theory that the number of primes in $P_n$ up to $x$ is $O(x/\log^2 x)$, with the constant certainly dependent on $n$.

Since this set has asymptotic density zero relative to the sequence of primes, there must be arbitrarily long clusters of primes that do not lie in $P_n$ (if there were no clusters of length $m$ then the lower relative density of $P_n$ would be at least $1/m$).