20120330, 10:34  #1 
Einyen
Dec 2003
Denmark
5627_{8} Posts 
Suggestion for new sieving software
Sascha77 came up with this problem/puzzle:
http://www.mersenneforum.org/showpos...7&postcount=74 Which corresponds to: Does any factor of a mersenne number 2np+1 have n=k*p. I checked all of GIMPS 33.5 million known factors and found 1 solution only: http://www.mersenneforum.org/showpos...0&postcount=81 http://www.mersenneforum.org/showpos...6&postcount=83 so this like a new kind of Wieferich prime, just as rare. So I wrote some programs and have been looking for other examples, but my programming skill is limited to C with GNU MP, so they are not super fast. It would be very nice with an efficient siever like newpgen/srsieve for numbers of the form 2*k*p^{2}+1, p prime (or k*p^{2}+1 whichever is easier) over k and p intervals. Maybe if people like Prime95/geoff/rogue/Jean PennĂ© who has the skills finds this interesting and have the time :) Last fiddled with by ATH on 20120330 at 10:40 
20120330, 11:19  #2 
"Bo Chen"
Oct 2005
Wuhan,China
7·23 Posts 
I think there should be infinite solutions to this problem, though it is rare.
For any given p, Consider the minimum number of A (A>1) that satisfy A^p = 1 (mod p^2) maybe help. 
20120330, 14:58  #3 
Romulan Interpreter
Jun 2011
Thailand
2^{2}·3·739 Posts 
There are certainly an infinite amount of solutions for composite p, and composite factors 2*k*p^2+1, as for example p=28, 56, 100, 126, 140, 240, 256, 280, 312, 336, 364, 512, 624, 748 etc.
The factors are prime for the trivial p=32, 64, 128, but also for other which are not powers of 2, like for example p=678: we have 2^6781 is divisible by the prime 10113049, which is 2*11*678^2+1. Contrary, there are not many solutions for odd composite p either. The only one I found 339: 2^3391 is divisible by the same prime, 10113049, which this time is 2*(11*4)*339^2+1, so k=44, considering that 678=2*339 (=2*3*113). But 2^1131 does not have any factor f whose f1 contain some square of 113. 
20120404, 13:03  #4  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
Code:
for(x=1,10000,for(k=1,x,if((2^x1)%(2*k*x^21)==0,print(k","x)))) Last fiddled with by science_man_88 on 20120404 at 13:08 

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