Quote:
Originally Posted by George M
I made a conjecture, but donâ€™t know how to prove it. Perhaps it could be related to Mersenne Primes?
Consider .
If then .
This is to say, if a natural number (positive integer) is equal to the product of the 1st prime, the 2nd prime, the 3rd prime, and so on, until the kth prime, then the amount of all the divisors of (including 1 and ) will be equal to .
How must we go about proving this?

By induction. If there are 2^k divisors for the product P, and you multiply P by some prime not in P (call it q), the divisors of the new number Pq are the divisors of P, together with q times the divisors of P. Since there is no overlap (why?), there are 2^k + 2^k = 2^(k+1) divisors. Now just test the base case and you can add QED.
Quote:
Originally Posted by George M
Perhaps it could be related to Mersenne Primes?

No.