Prime Number Theory says that the chance of a random integer x being prime is about 1/log x
For x = n!+/-1, chance is about w/log n!
To find the probability over a given range we integrate from n = A to n = B.
You can use approximation to Stirling's Formula, where n! ~= n^n * e^-n and some log identities, i.e.
1/log(n^n * e^-n)
= 1/(log(n^n) + log(e^-n))
= 1/(n*log(n) - n*log(e))
= 1/(n*log(n) - n)
Integrating 1/(n*log(n) - n) gives log(log(n)-1) so:
Number of Primes = w * (log(log(B)-1) - log(log(A)-1))
Weight is scaled so w=1.0 has the same density of primes as a randomly chosen set of odd numbers of the same magnitude.
You can use fpsieve to find P1e6; the number of candidates remaining after testing the range n=100001 to 110000 for factors up to 1 million.
>fpsieve -n 100001 -N 110000 -P 1e6 -x
Running this I found:
2528 terms with factors for n!-1, so 7472 remain and
2651 terms with factors for n!+1, so 7349 remain.
That's a good start. I'll finish trying to crack it later, unless someone else wants to take on the challenge.