It comes from the original problem which renewed the motivation to find consecutives primes in arithmetic progression, namely:
```   Problem A:
Let p(n) denote the n-th prime, with p(1)=2, and let p(n)=a(n)*n+r(n)
be the Euclidean division of p(n) by n, with 0<=r(n)<=n-1.
Find k consecutive primes p(n), p(n+1), ..., p(n+k-1) with the same
rest, i.e. r(n)=r(n+1)=...=r(n+k-1).
```
Problem A was posed by A. Vavoda in the September issue of "Pour la Science", the french equivalent of "Scientific American". With Nik Lygeros and Michel Mizony from University Claude Bernard in Lyon, we proved that:
```(1) necessarily a(n)=a(n+1)=...=a(n+k-1)
(2) if we denote this common quotient by a, then a must be divisible by
each prime less or equal to k
```
We also found several solutions to this problem for a=6, 12, 18, 24 (k=3 or 4), and a=30 (k=3, 4, 5 or 6). For k=7 to 10, (2) states that we have to go to a=210. Then we discover that Problem A reduces to find k consecutive primes in arithmetic progression in a given domain, namely with floor(p(i)/i)=a for each i=n, n+1, ..., n+k-1. Using the inequalities
```   n*(ln(n)+ln(ln(n))-1) <= p(n) <= n*(ln(n)+ln(ln(n))-0.9484)
```
proved by Pierre Dusart for n>=39017 (to appear in Math. of Comp.), this gave us a "window" for a=210 which is approximately 0.4347341756*10^92 <= p <= 1.181702821*10^92. With the given x and m values, this gives the above range for N.