20190627, 10:11  #23 
Banned
"Luigi"
Aug 2002
Team Italia
3·1,609 Posts 

20190627, 13:36  #24 
Mar 2019
11000110_{2} Posts 
Perhaps the author of mmff could comment on the feasibility of modifying this program to find Double Wagstaff factors, and if so, how it would need to be changed?

20190627, 15:06  #25  
Sep 2003
101000011001_{2} Posts 
Quote:
It is not nearly as simple as the changes that were made to mfaktc to handle Wagstaff, which was a mere handful of #ifdef statements. The mfaktc code can handle any arbitrary exponent up to a certain limit, including MM31 (exponent = about 2.1 billion), but not the huge "double" exponents. I used it to test WW31 (exponent = about 715 million) up to TF=2^80 and found one small factor of about 62 bits. WW43 (exponent = about 2.9 trillion) is out of reach. The Double Mersenne search hasn't had any results beyond MM31. The most recent of four factors of MM31 was found in 2005, so probably mmff was not used to find it. Since Double Mersenne hasn't had much success and seems to have only two currently active users, and Wagstaff has much less overall interest than Mersenne, and creating Double Wagstaff code would involve nontrivial specialized effort, it's kind of hard to make a case for doing Double Wagstaff. 

20191115, 04:55  #26 
"Dylan"
Mar 2017
2·293 Posts 
I don't know if there has been any update to searching for factors of double Wagstaffs, but I searched for potential factors of W(W(42737)), using pfgw64 with the f1 flag, and the following input:
Code:
ABC2 2*$a*((2^42737+1)/3)+1 a: from 1 to 50000 Code:
2*21085*((2^42737+1)/3)+1 2*42289*((2^42737+1)/3)+1 Although this is reasonably quick (a test takes about half a second), there's a lot of candidates with a factor of 3, so a sieve would be good for a serious search. 
20191115, 06:11  #27 
Jun 2003
141D_{16} Posts 
Wagstaff factors must be 1 or 3 (mod 8). Your numbers are 7 (mod 8). So they can't divide.
But if you want to test them, you can use Pari/GP Code:
? W=(2^42737+1)/3; ? N=2*21085*((2^42737+1)/3)+1; ? Mod(2,N)^W+1 == 0 %1 = 0 Last fiddled with by axn on 20191115 at 06:19 
20191117, 21:12  #28 
"Dylan"
Mar 2017
586_{10} Posts 
Thanks axn for the advice. I've updated the ABC2 header:
Code:
ABC2 2*(4*$a)*((2^42737+1)/3)+1  2*(4*$a+3)*((2^42737+1)/3)+1 a: from 0 to 62500 Code:
2*(4*13918+3)*((2^42737+1)/3)+1 2*(4*18327+3)*((2^42737+1)/3)+1 2*(4*26962+3)*((2^42737+1)/3)+1 2*(4*34111)*((2^42737+1)/3)+1 2*(4*38520)*((2^42737+1)/3)+1 2*(4*45025)*((2^42737+1)/3)+1 2*(4*54945)*((2^42737+1)/3)+1 2*(4*58207)*((2^42737+1)/3)+1 2*(4*58308+3)*((2^42737+1)/3)+1 
20191118, 02:49  #29 
Jun 2003
19×271 Posts 
I did a search of all k < 10^6. Here are the primes. No factors, natch.
Code:
2*55675*((2^42737+1)/3)+1 2*73311*((2^42737+1)/3)+1 2*107851*((2^42737+1)/3)+1 2*136444*((2^42737+1)/3)+1 2*154080*((2^42737+1)/3)+1 2*180100*((2^42737+1)/3)+1 2*219780*((2^42737+1)/3)+1 2*232828*((2^42737+1)/3)+1 2*233235*((2^42737+1)/3)+1 2*266260*((2^42737+1)/3)+1 2*284271*((2^42737+1)/3)+1 2*308163*((2^42737+1)/3)+1 2*319195*((2^42737+1)/3)+1 2*340684*((2^42737+1)/3)+1 2*341500*((2^42737+1)/3)+1 2*348316*((2^42737+1)/3)+1 2*367168*((2^42737+1)/3)+1 2*368020*((2^42737+1)/3)+1 2*406035*((2^42737+1)/3)+1 2*427771*((2^42737+1)/3)+1 2*464971*((2^42737+1)/3)+1 2*468720*((2^42737+1)/3)+1 2*470836*((2^42737+1)/3)+1 2*534735*((2^42737+1)/3)+1 2*535951*((2^42737+1)/3)+1 2*574444*((2^42737+1)/3)+1 2*576928*((2^42737+1)/3)+1 2*580056*((2^42737+1)/3)+1 2*623875*((2^42737+1)/3)+1 2*628611*((2^42737+1)/3)+1 2*639783*((2^42737+1)/3)+1 2*652996*((2^42737+1)/3)+1 2*691695*((2^42737+1)/3)+1 2*759304*((2^42737+1)/3)+1 2*771864*((2^42737+1)/3)+1 2*795195*((2^42737+1)/3)+1 2*813655*((2^42737+1)/3)+1 2*823683*((2^42737+1)/3)+1 2*902104*((2^42737+1)/3)+1 2*925608*((2^42737+1)/3)+1 2*953796*((2^42737+1)/3)+1 2*995536*((2^42737+1)/3)+1 
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