p(i)# divides n!
- p(i)# | (n+1)!
where p(i) is the largest prime <= n! and p(i)# is primorial.
So a gap same size n can be found at (n+1)!/p(i)#.
This helps to find g to be smaller, but p(i)# is still huge.
Also, by the same logic as with n! can be applied to p(i)# to have a gap
after p(i)#+1 of minimum size p(i)+1.
John W. Nicholson mailto:JohnW.Nicholson@...