20180102, 05:55  #1 
Dec 2017
2×5^{2} 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 . 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? If we do, perhaps we could build an algorithm to find a prime value for k such that is prime. 
20180102, 06:28  #2 
May 2007
Kansas; USA
2×5,153 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 20180102 at 06:41 
20180102, 06:44  #3  
Aug 2006
1011101010100_{2} Posts 
Quote:
No. 

20180102, 08:11  #4 
Dec 2017
2×5^{2} Posts 
I didn’t mean to put this thread there so sorry about that, but thank you for moving it :)

20180102, 08:13  #5  
Dec 2017
2·5^{2} Posts 
Quote:


20180102, 11:11  #6  
Feb 2017
A5_{16} Posts 
Quote:
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) 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; (2^11)*[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 (2^91)*[2^(91)] = 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 2^91 (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. Caveat: I am not sure if anybody has already stated any of the above, bar of course the consequences flowing from the EuclidEuler theorem Last fiddled with by gophne on 20180102 at 11:13 Reason: spelling/typo's 

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