... In general, finding a square root modulo a composite, m, is as difficult as factorizing m. If someone could do one of these things in polynomial time, they...
... There has been a delay since this was posted. Regarding the Sierpinski case: One such covering set is [13,5,7,41,73,17,193,6481,97,577] which have ...
... I translated it to GP. Here is a typical random run with a small composite modulus: No sqrt of Mod(2437846,10097063) because of loop. No sqrt of...
... Defined it in a different way, so that each of the numbers in the series must be multiplied by all primes with multiplicative order base b of 1 to get to a...
Hi, All LLR/LLRNET users, I wish to remind you of the real features of these programs : While testing k*b^n+1 or k*b^n-1 candidates, LLRNET/LLR can only prove ...
... Thanks David. I averaged the step length of the 8 successful runs you did, to get an average of 1700, which is slightly more than half the square root of...
... For the case r=2 and k=1, see Cohen, CCANT, Algorithm 1.5.1. As Phil remarked, there are better methods when p != 1 mod 8, or p = 1 mod 2^e with large e. ...
Hello all, can everbody make a PRP-test with PFGW.exe (PFGW Version 20020515) for number "3*20000#-1" ? I don't care number is prime or not. I need the time in...
Big congratulations from me to your new AP record ! Norman ... ___________________________________________________________ Telefonate ohne weitere Kosten vom...
... Indeed. But that fact is not appreciated here: http://mersennewiki.org/index.php/Sierpinski/Riesel_Base_5 ... Maybe George should append a health warning...
Hello, I would like to know if anyone know how to set the range of one variable based on another variable. I tried to find generalized cullen and woodall...
I suggest that you read the ABC format document. Example, for generalized woodall, ABC2 ($a+$b)*$a^($a+$b)-1 a: from 3 to 100 b: from 1 to 100 If you are...
I just heard a remarkable radio programme on the BBC world service: http://www.bbc.co.uk/worldservice/programmes/discovery.shtml ... wherein Greg Chaitin...
Hi, I have been looking at PRP's of the form 109999...91,109999...93,109999...97 using ABC2 10*(10*10^$a+10^$a-1)+m a: from 1 to 5000 where m=1,3,7 Some...
... Thanks for that, David. Omega is very interesting. http://www.youtube.com/results?search_query=Gregory+Chaitin&search=Search might be entertaining for...
... The form 11*10^n-k with k=3,7,9 requires ECPP for n=O(5000). You might obtain a Primo proof for 11*10^4983-9 is Fermat and Lucas PRP! in (very) roughly 3...
I know this question has been asked before, but has any progress been made on getting a buildable source for the Linux version of PFGW? I recently got 1000...
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