# Prime chains with large gaps

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 ?

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 $$1/m$$).