Why does the primorial $23\#$ come up so often in long prime arithmetic progressions? This section of the Wikipedia article on the Green-Tao theorem gives examples of the longest known arithmetic progressions of prime numbers. For every known arithmetic progression of $24$ or more consecutive primes, the common difference is a multiple of the primorial $23\# = 223,092,870$ (the product of all primes up to $23$).
Is it a coincidence that $23$ is the largest prime less than the length of the long sequences? With enough computing power, would we eventually expect to find arithmetic sequences of $29$ or more primes whose common differences are multiples of $29\#$? If not, is there something special about the number $23\#$, or does it simply reflect the limits of our computing power?
 A: Suppose you want to construct an arithmetic sequence of six primes and the difference is not a multiple of $5$. You start with, let us say, $7$ and proceed with a difference of $12$ to generate $7, 19, 31, 43, 55, 67.$ Uh-oh, you got a multiple of $5$.
That's because when the difference between successive terms is not a multiple of $5$, the sequence modulo $5$ will inevitably cycle through all residues modulo $5$ and thus you inevitably hit zero modulo $5.$
You could avoid this by using $5$ itself as the multiple of $5$, but then $5$ has to begin the sequence and you find the sixth term is a multiple of $5$ again, this time composite. With a difference of $12$ like before, you get $5, 17, 29, 41, 53, 65.$ You started with $5$ and then made five increments with a common difference of $12$, so could not avoid another multiple of $5.$
You get similar problems with differences that are not multiples of $2$ or $3.$ To get a sequence of primes as long as six terms, the difference must be divisible by all of $2, 3, 5$ and thus divisible by $5$# $=30.$ The sequence $7, 37, 67, 97, 127, 157$ makes it with a difference of exactly $30.$
Now try this logic with a sequence that's $24$ primes long and infer the difference between successively terms has to have all primes factors up to and including $23.$
A: Given an arithmetic progression $a,a+d,\ldots, a+(n-1)d$ and a prime $p\le n$ with $p\nmid d$, at least one of the $a+kd$ will be a multiple of $p$.
