20151026, 01:27  #45 
"Forget I exist"
Jul 2009
Dumbassville
10000010110001_{2} Posts 
the reason I asked it because they very slightly mentioned something in PM that was supposed to speed it up at very least.

20151026, 01:31  #46  
"Rashid Naimi"
Oct 2015
Remote to Here/There
1,931 Posts 
Quote:
I don't think it is necessarily worse than trial division in general though. There should be a medium range where less trials than trial division would result in primes specially if you write the routine intelligently by keeping the sums as low as possible but for very large primes the sums will diverge very rapidly. You don't really need to calculate the base primes either. you can use the theorem to come up with solutions to that such as using double factorials of a composite odd integer, rather than primes multiplied. the double factorial can be calculated using product of sequence formula. To further keep the sum from diverging you can factor out large number of known primes. But all this would not do for a billion plus digits. As I mentioned before there are better approaches to this. BTW, thank you for being one of the few intelligent posters to this thread. 

20151026, 01:34  #47 
"Rashid Naimi"
Oct 2015
Remote to Here/There
1,931 Posts 
Please read the red comment in my earlier post. I thought it was distinct enough. Obviously I was wrong.

20151026, 02:05  #48 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}·7·163 Posts 

20151026, 02:31  #49 
"Rashid Naimi"
Oct 2015
Remote to Here/There
1,931 Posts 

20151026, 03:07  #50 
Jun 2003
4,721 Posts 

20151026, 05:42  #51 
"Dana Jacobsen"
Feb 2011
Bangkok, TH
2×3×151 Posts 
This makes me think of Maurer or ShaweTaylor methods for generating proven primes. There we get to double or triple the bits every round. One could do something similar with other BLS75 theorems (Maurer's algorithm can be seen as a case of theorem 3) or even ECPP. Just build up the proof backwards to get larger and larger numbers. It makes nice large proven primes, but I seriously doubt it would scale well to millions of digits.

20151026, 07:28  #52  
∂^{2}ω=0
Sep 2002
República de California
2^{2}·31·79 Posts 
Quote:
Counter #1: A = {2,3}, B = 2, C = 3. Let b = 2^1, c = 3^3. Then b+c = 29 is prime, but b+c = 25 is not. Counter #2: A = {2,3,5,7,11}, B = 2, C = {3,5,7,11}. Let b = 2^1, c = 3*5*7*11 = 1155. Then b+c = 1153 is prime, but b+c = 1157 = 13*89 is not. 

20151026, 07:39  #53  
Romulan Interpreter
Jun 2011
Thailand
2^{2}·7·317 Posts 
Quote:


20151026, 07:48  #54 
Jun 2003
4,721 Posts 

20151026, 09:44  #55  
Nov 2003
2^{6}×113 Posts 
Quote:
is hardly new. See the paper "Primes at a Glance" by John Selfridge et.al. 

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