So how must we be able to prove the following?
I made a conjecture, but don’t know how to prove it. Perhaps it could be related to Mersenne Primes?
Consider [TEX]A =\{all \ divisors \ of \ x \in \mathbb{N}\}[/TEX]. If [TEX]x = \prod_{m=1}^{k}p_m[/TEX] then [TEX]n(A) = 2^k[/TEX]. This is to say, if a natural number (positive integer) [TEX]x[/TEX] 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 [TEX]x[/TEX] (including 1 and [TEX]x[/TEX]) will be equal to [TEX]2^k[/TEX]. How must we go about proving this? If we do, perhaps we could build an algorithm to find a prime value for k such that [TEX]2^k  1[/TEX] is prime. 
This thread was originally in the Riesel and Sierpinski conjectures project (CRUS). I have moved it to Miscellaneous Math. If one of the supermods feels that it should be moved somewhere else, please feel free.

[QUOTE=George M;475949]I made a conjecture, but don’t know how to prove it. Perhaps it could be related to Mersenne Primes?
Consider [TEX]A =\{all \ divisors \ of \ x \in \mathbb{N}\}[/TEX]. If [TEX]x = \prod_{m=1}^{k}p_m[/TEX] then [TEX]n(A) = 2^k[/TEX]. This is to say, if a natural number (positive integer) [TEX]x[/TEX] 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 [TEX]x[/TEX] (including 1 and [TEX]x[/TEX]) will be equal to [TEX]2^k[/TEX]. How must we go about proving this?[/QUOTE] 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=George M;475949]Perhaps it could be related to Mersenne Primes?[/QUOTE] No. 
[QUOTE=gd_barnes;475955]This thread was originally in the Riesel and Sierpinski conjectures project (CRUS). I have moved it to Miscellaneous Math. If one of the supermods feels that it should be moved somewhere else, please feel free.[/QUOTE]
I didn’t mean to put this thread there so sorry about that, but thank you for moving it :) 
[QUOTE=CRGreathouse;475957]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.
No.[/QUOTE] I was able to use induction? I didn’t think that I could use induction on an equation like this. It was simpler than I thought. Well, thank you very much! I perhaps should have posted this on the Mathematics Stack Exchange but figured it was related to mersenne primes since we have prime numbers and 2^k. 
[QUOTE=George M;475949]I made a conjecture, but don’t know how to prove it. Perhaps it could be related to Mersenne Primes?
Consider [TEX]A =\{all \ divisors \ of \ x \in \mathbb{N}\}[/TEX]. If [TEX]x = \prod_{m=1}^{k}p_m[/TEX] then [TEX]n(A) = 2^k[/TEX]. This is to say, if a natural number (positive integer) [TEX]x[/TEX] 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 [TEX]x[/TEX] (including 1 and [TEX]x[/TEX]) will be equal to [TEX]2^k[/TEX]. How must we go about proving this? If we do, perhaps we could build an algorithm to find a prime value for k such that [TEX]2^k  1[/TEX] is prime.[/QUOTE] Hi George M When I was working/playing around with "perfect even numbers" related to mersenne numbers and Euclid's related proof  that when 2^k1 is prime, the equation 2^(k1)*(2^k1) would produce an (even) [B]perfect number[/B], e.g for M5, this would give the perfect number (31)*(16)=496 [Euler proved the converse...that all perfect numbers have that form.....source [url]https://primes.utm.edu/notes/proofs/EvenPerfect.html[/url] ] Definition of a perfect number being......[url]https://en.wikipedia.org/wiki/Perfect_number[/url] Breaking this up a bit, I tabulated the following listing of this equation of Euclid, for all mersenne (odd) numbers to any selected odd number; (2^11)*[2^(11)] = 00001* 00001 = 00001 (2^21)*[2^(21)] = 00003* 00002 = 00006.....prf factors(006) 1,2  3,6...................004 terms ~ 2x n (2^31)*[2^(31)] = 00007* 00004 = 00028.....prf factors(028) 1,2,4  7,14,28.........006 t (2^51)*[2^(51)] = 00031* 00016 = 00496.....prf factors(496) 1,2,4,8,16  31,62,124,248,496..010 t (2^71)*[2^(71)] = 00127* 00064 = 08128.....prf factors(8128)1,2,4,8,16,32,64  127,254,508,1016,2032,4064,8128................014 t (2^[COLOR="red"]9[/COLOR]1)*[2^([COLOR="red"]9[/COLOR]1)] = [COLOR="Red"]00511[/COLOR]* 00256 = 130816...prf factors(1308168128) ................<>018 t, however, ignoring the additional factors introduced by the fact that "511" is not prime, this formulaic expansion would allways produce a perfect number! bar the additional factors introduced by the fact that 2^91 (511) is prime...I think this was the essence of Euclids proof? In the tabulation, the first string of factors are the factors of (2^k1) and the second string of factors are the first set of factors multiplied by the mersenne number ~ 2^k1, bar when 2^k1 is not prime. Interestingly, the first set of factors (bar additional factors introduced when 2^k1<>prime) adds up to 2^k1, and the sum of the second set of factors (bar additional factors when 2^k1<>prime) = (2^k1)^2 Not sure if the above has any bearing on your conjecture. [I] Caveat: I am not sure if anybody has already stated any of the above, bar of course the consequences flowing from the EuclidEuler theorem[/I] 
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