20211126, 22:14  #1 
Feb 2004
France
3^{2}×103 Posts 
Universal seeds of the LLT for Mersenne and Wagstaff numbers
You probably know that there are 3 Universal Seeds for starting the LLT for Mersenne numbers: 4, 10, and 2/3.
It appears that 2/3 modulo Mq = (2^q+1)/3 = Wq, a Wagstaff number. Also, you may know that there is a conjecture about proving that a Wagstaff number Wq is prime by using a cycle of the DiGraph under x^22 modulo Wq. The seed we found years ago is : 1/4. And kijinSeija found that 1/4 = W(q2). Moreover, I've found that 1/10 and 1/Mq seem also to be usable as Universal Seed for this conjecture, building a new bridge between Mersenne and Wagstaff primes, since the seeds of the Wagstaff conjecture are inverse modulo Wq of the Universal Seeds of the Mersenne LLT. Pari/gp: T(q)={M=2^q1;W=(2^q+1)/3;S0=Mod(1/4,W);S=S0;print(S);for(i=1,q+1,S=Mod(S^22,W);print(S))} Try and replace 1/4 by 1/10 and 1/Mq. It's perfectly clear for 1/4 and 1/10. It's a little bit different for 1/Mq. It seems to depend if q=4k+1 or 4k1. It needs deeper study. 
20211126, 22:38  #2  
Sep 2002
Database er0rr
4145_{10} Posts 
Quote:
Code:
wag(q)=W=(2^q+1)/3;S0=S=Mod(3/2,W);for(i=2,q,S=S^22);S==S0; Are 4, 10 and 2/3 the only ones for Mersenne? I remember Lehmer had said something about these. Last fiddled with by paulunderwood on 20211126 at 22:48 

20211126, 23:49  #3 
Sep 2002
Database er0rr
5·829 Posts 
More on the seed 3/2 for Wagstaff numbers.
Mod(Mod(x,W),x^23/2*x+1) is at the heart because S = S0 = 3/2 = x+1/x which leads to the recurrence S=S^22 mod W The solution for x is ( 3/2 + sqrt( (3/2)^24 ) / 2 = ( 3 + 2 * sqrt(2) ) / 4 That is 4*x  3 == + 2 * sqrt(2) So we can use a power of Mod(Mod(4*x3,W),x^28). By trial and error Mod(Mod(4*x3,W),x^28)^(W+1)+119 == 0 Edit. That marked red is wrong, but somehow the result is good! Stumped! Last fiddled with by paulunderwood on 20211127 at 01:30 
20211127, 00:35  #4  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
1100101011110_{2} Posts 
Quote:
The ninth page of Bas Jensen's PhD number theory thesis https://scholarlypublications.univer...3A2919366/view makes reference to a 1996 conjecture by G. Woltman, proven 4 years later. Last fiddled with by kriesel on 20211127 at 00:37 

20211127, 05:08  #5  
Einyen
Dec 2003
Denmark
3,313 Posts 
Quote:
https://www.mersenneforum.org/showpo...0&postcount=46 A_{0}=A_{1}=4, A_{N}=14*A_{N1}A_{N2} B_{0}=B_{1}=10, B_{N}=98*B_{N1}B_{N2} Last fiddled with by ATH on 20211127 at 05:11 

20211127, 08:52  #6  
Feb 2004
France
3^{2}·103 Posts 
Quote:
I see that 2/3 = Wq was already known. 

20211127, 08:54  #7 
Feb 2004
France
3^{2}·103 Posts 

20211127, 09:22  #8 
Sep 2002
Database er0rr
4145_{10} Posts 
It is not so great since (3/2)^22 = 1/4 which is already known.
Just as (3)^((Mp  1)/2) == 1 mod Mp for Mersenne primes, we have (7)^((Wq  1)/2) == 1 mod Wq for Wagstaff PRPs. The latter can be 7^((Wq  1 )/2) == 1 mod Wq. Last fiddled with by paulunderwood on 20211127 at 10:09 
20211127, 13:35  #9 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2·17·191 Posts 

20211130, 22:15  #10 
Feb 2004
France
1110011111_{2} Posts 
Hello.
I've spent some time searching for Universal Seeds for the LLTlike test for Wagstaff numbers. Here attached is a first set of findings. Everything has been checked for all known Wagstaff primes and for Wagstaff notprimes up to q = 10,000. It may be strange, but 23/8 mod Wq is a Universal seed! And 1/10 is, sometimes... Enjoy Tony 
20211208, 01:15  #11 
Bemusing Prompter
"Danny"
Dec 2002
California
2×3^{2}×137 Posts 
Probably a stupid question, but my knowledge is not up to scratch.
It is my understanding that "2/3" is shorthand for (2 mod M(p))(3 mod M(p))^{1} = W(p). I was indeed able to confirm that W(p) works as a seed value when testing whether M(p) is prime. However, it's not clear how (2 mod M(p))(3 mod M(p))^{1} evaluates to a Wagstaff number. I can see where the "2/3" comes from because 2 mod M(p) = 2 and 3 mod M(p) = 3 for p > 2. But it doesn't make sense to use a fraction as a seed value. Or does (3 mod M(p))^{1} mean something other than 1 / (3 mod M(p)) in this case? 
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