20180502, 18:54  #12 
Sep 2002
Database er0rr
5·29·31 Posts 

20180502, 18:58  #13 
Dec 2017
2^{4}×3×5 Posts 

20180502, 20:03  #14 
Sep 2002
Database er0rr
5·29·31 Posts 
I now have a new test to add to the previous 2:
Code:
f(p)=mp=2^p1;a=Mod(2,mp);for(b=1,p1,a=a/2+1/a);a==2 f(p)=mp=2^p1;a=Mod(2,mp);for(b=1,p1,a=a/2+3/a);a==0 f(p)=mp=2^p1;a=Mod(2,mp);for(b=1,p1,a=a/21/a);a==0 [new] 
20180502, 23:06  #15 
Einyen
Dec 2003
Denmark
2·3^{2}·191 Posts 
I do not think I understand them, I do not use Pari/GP.
f(p)=mp=2^p1;a=Mod(2,mp);for(b=1,p1,a=a/2+1/a);a==2 You start with a=2 and then you to (p1) steps: a=(a/2 + 1/a)%Mp where 1/a is the modular inverse mod Mp, and the result after (p1) steps a==2 for Mersenne primes? What about a/2 when a is odd? is that integer division? 
20180502, 23:16  #16  
"Forget I exist"
Jul 2009
Dartmouth NS
10000011100010_{2} Posts 
Quote:


20180502, 23:31  #17  
Sep 2002
Database er0rr
10617_{8} Posts 
Quote:
Inverses are usually computed by the "extended Euclidean algorithm" which really slows down "f". Last fiddled with by paulunderwood on 20180502 at 23:47 

20180502, 23:43  #18  
Sep 2002
Database er0rr
118F_{16} Posts 
Quote:
f(p)=wp=2^p+1;a=Mod(2,wp);for(b=1,p1,a=a/21/a);a==1a==2 (You have to fiddle with Mod and lift to get a good result for W_{3}.) I have not tested this very far. Last fiddled with by paulunderwood on 20180503 at 00:09 

20180503, 00:05  #19  
"Forget I exist"
Jul 2009
Dartmouth NS
20E2_{16} Posts 
Quote:
.......... 2 3 1 4 5 3 6 3 2 3 In other words. Last fiddled with by science_man_88 on 20180503 at 00:07 

20180503, 00:58  #20  
Einyen
Dec 2003
Denmark
2×3^{2}×191 Posts 
Quote:
So testing all 3 algorithms for all primes up to 11K they finds all Mersenne Primes up to 11213, but that does not prove them. Though it would be weird for them to find so many Mersenne Primes and then fail for higher primes. Last fiddled with by ATH on 20180503 at 00:59 

20180503, 02:56  #21  
Sep 2002
Database er0rr
10617_{8} Posts 
Quote:
Or if a is even ">>1", else "<<t", but testing for even is slow in Parigp. The hard thing remains computing of 1/s (mod Mp). Last fiddled with by paulunderwood on 20180503 at 03:05 

20180504, 11:36  #22 
Sep 2002
Database er0rr
5·29·31 Posts 
Here is another test for Wagstaff (2^p+1)/3:
Code:
f(p)=wp=2^p+1;a=Mod(1/2,wp);for(b=1,p1,a=a/21/a);lift(a*2)%(wp/3)==1 Last fiddled with by paulunderwood on 20180504 at 11:40 
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