20160608, 01:20  #1 
Sep 2002
Database er0rr
3206_{10} Posts 
Alternative to LL
Code:
f(p)=mp=2^p1;a=Mod(2,mp);for(b=1,p1,a=(a/2+1/a));a==2 
20160608, 02:03  #2 
Romulan Interpreter
Jun 2011
Thailand
20530_{8} Posts 

20171113, 23:49  #3  
Sep 2002
Database er0rr
2·7·229 Posts 
Quote:
Code:
f(p)=mp=2^p1;a=Mod(2,mp);for(b=1,p1,a=(a/2+3/a));a==0 Last fiddled with by paulunderwood on 20171113 at 23:56 

20171113, 23:56  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
comes down to where the squares are on the arithmetic progression y(mp)6. parity of x is the parity of y.
Last fiddled with by science_man_88 on 20171113 at 23:57 
20171211, 00:56  #5 
Sep 2002
Database er0rr
2·7·229 Posts 
Here is another test for Mersennes:
Code:
f(p)=local(mp=2^p1,a=Mod(2,mp),b);for(b=3,p,a=a^2*21);a==0 
20171211, 01:05  #6  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20171211 at 01:08 

20171211, 02:54  #7 
"Sam"
Nov 2016
116_{16} Posts 
I don't know if this is useful but https://oeis.org/A002812 happens to be the subset of the sequence https://oeis.org/A110293. Like the Mersenne sequence 2^n1, this is also a divisibility sequence.
a(n) = 2*a(n1)^2  1, starting a(0)=n b(n) = b(n1)^22, starting b(0)=n are generalizations of https://oeis.org/A110293 I don't know the sequences which a(n) or b(n) are subsets of. That is, there exists sequences A(n) and B(n) such that A(2^n) = a(n) B(2^n) = b(n) The sequences a(n) and b(n) For further discussion let ll(2^n) be the sequence in https://oeis.org/A002812 and ll(n) be the sequence in https://oeis.org/A110293 Another question that comes up is how are ll(n) and 2^n1 related to eachother other than the fact that if 2^n1 is prime, then 2^n1 divides ll(2^(n2)). 
20180502, 15:37  #8  
Sep 2002
Database er0rr
6206_{8} Posts 
Quote:
1^2+6 == 0 mod 2^31 5^2+6 == 0 mod 2^51 11^2+6 == 0 mod 2^71 1405^2+6 == 0 mod 2^131 Last fiddled with by paulunderwood on 20180502 at 15:40 

20180502, 17:32  #9 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
12²+6== 0 mod 2^41
Last fiddled with by science_man_88 on 20180502 at 17:32 
20180502, 17:57  #10  
Feb 2017
Nowhere
CB3_{16} Posts 
Quote:
;) Last fiddled with by Dr Sardonicus on 20180502 at 18:05 

20180502, 18:40  #11 
Dec 2017
107 Posts 
Paul can you elaborate on this concept of rapid verification of Mprime?.

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