20050512, 02:02  #12  
Nov 2003
2^{2}×5×373 Posts 
Quote:
faster with SNFS or GNFS. SNFS lets us take advantage of the algebraic factor 2^152+1, but requires a quartic, which is suboptimal for numbers this size. 

20050512, 09:00  #13  
Apr 2004
Copenhagen, Denmark
2^{2}×29 Posts 
Quote:
Code:
Each lines contains the values: nbit nprimes a5max normmax normmax1 normmax2 murphye e ... 350 4 100000 6e15 2e14 2e12 1e09 1.69e09 360 4 200000 1.5e16 5e14 5e12 7.5e10 1.13e09 370 4 200000 4e16 1.5e15 1.5e13 5e10 7.75e10 380 4 500000 8e16 4e15 3e13 3.5e10 5.02e10 390 4 500000 3e17 1e16 1e14 2.5e10 3.53e10 400 4 500000 9e17 3e16 2.5e14 2e10 2.89e10 410 4/5 2000000 3e18 1e17 7e14 1.4e10 1.71e10 420 5 2000000 1e19 3e17 2e15 9e11 1.10e10 430 5 2000000 3e19 9e17 5e15 5e11 7.12e11 440 5 2000000 1e20 2.5e18 1.5e16 3e11 5.26e11 450 5 2000000 4e20 7e18 4e16 2e11 3.16e11 460 6 10000000 1e21 2e19 1e17 1.8e11 2.14e11 470 6 10000000 4e21 6e19 3e17 1e11 1.63e11 480 6 10000000 1.5e22 2e20 9e17 7e12 9.14e12 490 ? ? 5e22 5e20 2.5e18 500 7 100000000 1.5e23 1.5e21 7e18 3e12 >3.44e12 510 ? ? 5e23 4e21 2e19 520 ? ? 2e24 1.5e22 5e19 530 ? ? 6e24 4e22 1.5e20 540 ? ? 2e25 1e23 4e20 Don, Please just assign an a5 range for me as you see fit. Currently I have some machines sieveing for 3,437, but I can start searching when they are done sieving.  Cheers, Jes Last fiddled with by JHansen on 20050512 at 09:02 

20050512, 19:30  #14  
Mar 2003
79 Posts 
Hi Bob,
Quote:
f(x) = x^2  x  1 g(x) = 2^152*x2^3041 x=(2^152+1/2^152) Clever selection of polynomials is certainly not my forte and I've probably missed something. Any tips? Here is PARI code to show the process I used to reach the quadratic and linear: Code:
x^608  x^456 + x^304  x^152 + 1 y=x^152 y^4  y^3 + y^2  y + 1 (y^4 + y^2)  (y^3 + y) + 1 (y^4 + y^2)  (y^3 + y) + 1 ( (y^4 + y^2)  (y^3 + y) + 1 ) / y^2 ( (y^2 + 1)  (y^1 + y^1) + y^2 ) y^2 + 1  y^1  y^1 + y^2 y^2 + y^2  (y^1 + y^1) + 1 z=y+y^1 y^2 + y^2  z + 1 z^2= (y^2 + y^2 + 2) z^2  2  z + 1 z^2  z  1 n=(2^760+1)/(2^152+1) f(x) = x^2  x  1 g(x) = 2^152*x2^3041 m=(2^152+1/2^152) f(m)%n g(m)%n 

20050512, 19:53  #15  
Nov 2003
2^{2}·5·373 Posts 
Quote:
2^760 + 1 = x^5 + 1 with x = 2^152. But x^5+1 = (x+1)(x^4x^3+x^2x+1) So just use x  M and x^4x^3+x^2x+1 with M = 2^152 It is best NOT to convert the reciprocal quartiic into a quadratic in (x+1/x) because doing so blows up the coefficients on the linear side by too much. 

20050513, 04:07  #16  
Mar 2003
New Zealand
13×89 Posts 
Quote:


20050513, 06:38  #17  
Jul 2003
So Cal
949_{16} Posts 
Quote:
Greg 

20050513, 08:25  #18  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2^{3}·5·283 Posts 
Quote:
So we won't get a >p100 penultimate factor any more, but it's still worth finishing with SNFS. Luckily the cofactor is still composite, or we would have spent quite some time having run the survey sieving to no avail. Paul 

20050513, 10:56  #19  
Nov 2003
1110100100100_{2} Posts 
Quote:
have a higher density of small primes that split completely. This would make norms slightly more likely to be smooth. 

20050705, 03:28  #20  
Jan 2004
7·19 Posts 
Quote:
Thanks. 

20090315, 04:22  #21 
Dec 2008
7^{2}×17 Posts 
Sorry to be offtopic, but you are so lucky that you sleep 7 hours. I normally get 35 hours of sleep.
Last fiddled with by flouran on 20090315 at 04:22 