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 2018-01-02, 05:55 #1 George M   Dec 2017 2×52 Posts 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 $A =\{all \ divisors \ of \ x \in \mathbb{N}\}$. If $x = \prod_{m=1}^{k}p_m$ then $n(A) = 2^k$. This is to say, if a natural number (positive integer) $x$ is equal to the product of the 1st prime, the 2nd prime, the 3rd prime, and so on, until the k-th prime, then the amount of all the divisors of $x$ (including 1 and $x$) will be equal to $2^k$. 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 $2^k - 1$ is prime.
 2018-01-02, 06:28 #2 gd_barnes     May 2007 Kansas; USA 23·1,301 Posts 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. Last fiddled with by gd_barnes on 2018-01-02 at 06:41
2018-01-02, 06:44   #3
CRGreathouse

Aug 2006

3·1,993 Posts

Quote:
 Originally Posted by George M I made a conjecture, but don’t know how to prove it. Perhaps it could be related to Mersenne Primes? Consider $A =\{all \ divisors \ of \ x \in \mathbb{N}\}$. If $x = \prod_{m=1}^{k}p_m$ then $n(A) = 2^k$. This is to say, if a natural number (positive integer) $x$ is equal to the product of the 1st prime, the 2nd prime, the 3rd prime, and so on, until the k-th prime, then the amount of all the divisors of $x$ (including 1 and $x$) will be equal to $2^k$. 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.

2018-01-02, 08:11   #4
George M

Dec 2017

3216 Posts

Quote:
 Originally Posted by gd_barnes 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.
I didn’t mean to put this thread there so sorry about that, but thank you for moving it :)

2018-01-02, 08:13   #5
George M

Dec 2017

2·52 Posts

Quote:
 Originally Posted by CRGreathouse 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.
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.

2018-01-02, 11:11   #6
gophne

Feb 2017

3·5·11 Posts

Quote:
 Originally Posted by George M I made a conjecture, but don’t know how to prove it. Perhaps it could be related to Mersenne Primes? Consider $A =\{all \ divisors \ of \ x \in \mathbb{N}\}$. If $x = \prod_{m=1}^{k}p_m$ then $n(A) = 2^k$. This is to say, if a natural number (positive integer) $x$ is equal to the product of the 1st prime, the 2nd prime, the 3rd prime, and so on, until the k-th prime, then the amount of all the divisors of $x$ (including 1 and $x$) will be equal to $2^k$. 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 $2^k - 1$ is prime.
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^k-1 is prime, the equation 2^(k-1)*(2^k-1) would produce an (even) perfect number, 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 https://primes.utm.edu/notes/proofs/EvenPerfect.html ]

Definition of a perfect number being......https://en.wikipedia.org/wiki/Perfect_number

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;

(2^1-1)*[2^(1-1)] = 00001* 00001 = 00001
(2^2-1)*[2^(2-1)] = 00003* 00002 = 00006.....prf factors(006) 1,2 -- 3,6...................004 terms ~ 2x n
(2^3-1)*[2^(3-1)] = 00007* 00004 = 00028.....prf factors(028) 1,2,4 -- 7,14,28.........006 t
(2^5-1)*[2^(5-1)] = 00031* 00016 = 00496.....prf factors(496) 1,2,4,8,16 -- 31,62,124,248,496..010 t
(2^7-1)*[2^(7-1)] = 00127* 00064 = 08128.....prf factors(8128)1,2,4,8,16,32,64 -- 127,254,508,1016,2032,4064,8128................014 t
(2^9-1)*[2^(9-1)] = 00511* 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 2^9-1 (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^k-1) and the second string of factors are the first set of factors multiplied by the mersenne number ~ 2^k-1, bar when 2^k-1 is not prime.

Interestingly, the first set of factors (bar additional factors introduced when 2^k-1<>prime) adds up to 2^k-1, and the sum of the second set of factors (bar additional factors when 2^k-1<>prime) = (2^k-1)^2

Not sure if the above has any bearing on your conjecture.

Caveat: I am not sure if anybody has already stated any of the above, bar of course the consequences flowing from the Euclid-Euler theorem

Last fiddled with by gophne on 2018-01-02 at 11:13 Reason: spelling/typo's

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