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.)

  • $\begingroup$ Erick, do you have a reference for the theorem of Shiu ? Thank you in advance. $\endgroup$ – Rodney Coleman Jan 5 '14 at 19:59
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    $\begingroup$ @RodneyColeman See D. K. L. Shiu, "Strings of Congruent Primes", J. London Math. Soc. (2000) 61 (2): 359--373. jlms.oxfordjournals.org/content/61/2/359.abstract $\endgroup$ – Erick Wong Jan 5 '14 at 20:09

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