20190626, 11:26  #12 
Sep 2003
5·11·47 Posts 

20190626, 13:37  #13  
Sep 2003
A19_{16} Posts 
Quote:
As you mentioned, with mfaktc.exe it suffices to set the DWAGSTAFF flag to make it find Wagstaff factors instead of Mersenne factors. So maybe that will work with mmff.exe as well, and it might be possible to find a largeish factor for W(W(43)). Edit: from looking at the source code, it's not that simple. Last fiddled with by GP2 on 20190626 at 14:14 

20190626, 17:07  #14  
"Dylan"
Mar 2017
1001001100_{2} Posts 
Quote:
Code:
ABCD 2*$a*((2^p+1)/3)+1 (Of course, more work is needed to get the sieve to work.) 

20190626, 17:41  #15  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
110010001100_{2} Posts 
Quote:
Code:
n Phi_n(2) known factors of (2^Phi_n(2)+1)/3 Last fiddled with by sweety439 on 20190626 at 17:46 

20190626, 18:39  #16 
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 
The common generalization could be:
Code:
a(n)=polcyclo(h*n,2) for fixed h>0 integer. For h=2 a(a(p))=W(W(p)) if p and W(p)=(2^p+1)/3 is prime. And you can see this for h>2 also. Or you can even drop the n=p requirement (ofcourse in this case a(n)!=M(n) for h=1 etc.), note that we can see a(n)=prime or a(a(n))=prime for composite n values also. 
20190626, 19:45  #17  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·11·73 Posts 
Quote:
However, there are no known n such that Phi(n,2) is composite but Phi(Phi(n,2),2) is prime, also no known n such that Phi(n,2) is composite but Phi(2*Phi(n,2),2) is prime Last fiddled with by sweety439 on 20190626 at 19:51 

20190626, 19:58  #18 
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 

20190626, 21:37  #19 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{5}·3·101 Posts 
Shhhhhh.... How dare you! You are arguing with the inventor of the Double Wagstaff primes. No one thought of that before, and now someone finally has.
{/sarcasm} 
20190627, 01:56  #20  
"Mark"
Apr 2003
Between here and the
2×7×467 Posts 
Quote:


20190627, 02:38  #21 
Jul 2003
2×307 Posts 
hi,
mmff does not (yet) run with nvidia turing cards 
20190627, 04:13  #22  
"Dylan"
Mar 2017
24C_{16} Posts 
Quote:
Exponents 31, 61, 89, 107, 127 for double Mersennes and Exponents <= 223 for Fermat numbers. Largest bit level it can test depends on the exponent (see lines 356455 of mfaktc.c in the source of mmff). 

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