Found a clever little algorithm for computing the product of all primes between n-m without recomputing them.

We'll start with the product of all primes up to some n.

so [2, 2*3, 2*3*5, 2*3*5*,7..] etc

prods = []
i = 0
total = 1
while i < 100:
....total = total*primes[i]
....prods.append(total)
....i = i + 1

Terrible variable names, can't be arsed at the moment.
The result is a list with the values

2, 6, 30, 210, 2310, 30030, etc.

Now assume you have two factors,with indexes i, and j, where j>i

You can calculate the gap between the two corresponding primes easily.
A gap is defined at the product of all primes that fall between the prime indexes i and j.

To calculate the gap between any two primes, merely look up their index, and then do..
prods[j-1]/prods[i]

That is the product of all primes between the J'th prime and the I'th prime

To get the product of all primes *under* i, you can simply look it up like so:

prods[i-1]

Incidentally, finding a number n that is equivalent to (prods[j+i]/prods[j-i]) for any *possible* value of j and i (regardless of whether you precomputed n from the list generator for prods, or simply iterated n=n+1 fashion), is equivalent to finding an algorithm for generating all prime numbers under n.

Hypothetically you could pick a number N out of a hat, thats a thousand digits long, and it happens to be the product of all primes underneath it.

You could then start generating primes by doing
i = 3
while i < N:
....if (N/k)%1 == 0:
........factors.append(N/k)
....i=i+1

The only caveat is that there should be more false solutions as real ones. In otherwords theres no telling if you found a solution N corresponding to some value of (prods[j+i]/prods[j-i]) without testing the primality of *all* values of k under N.