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@...

http://web2.airmail.net/johnwn1/