20110822, 16:28  #1 
(loop (#_fork))
Feb 2006
Cambridge, England
2^{3}·797 Posts 
SNFS polynomials via lindep
Also, it's a Fibonacci number so it's SNFSable (play around with lattice reduction of {FibN, fib(A)^6, fib(A)^5*fib(B) ... } for A,B near N/6).
Code:
A=fibonacci(1049) u=fibonacci(174) v=fibonacci(175) ? lindep([A,u^6,u^5*v,u^4*v^2,u^3*v^3,u^2*v^4,u*v^5,v^6]) %14 = [1, 1, 0, 15, 20, 30, 18, 5]~ Code:
n: 11306054131787655968747547768061150919509803257954523210546639678264631549875906520129375307461813768927343896318137750949420159685553746180237040613218941447819843916894065106795775853 Y1: 1033628323428189498226463595560281832 Y0: 1672445759041379840132227567949787325 skew: 0.76 c6: 5 c5: 18 c4: 30 c3: 20 c2: 15 c0: 1 lpbr: 29 lpba: 29 mfbr: 58 mfba: 58 alambda: 2.6 rlambda: 2.6 alim: 42000000 rlim: 42000000 Last fiddled with by fivemack on 20110822 at 16:31 
20110822, 16:35  #2 
"Ben"
Feb 2007
2^{3}×419 Posts 
Ahh, I didn't know what number he was attempting, good catch!
@LaurV: you're a newcomer so it may not be obvious that this number had special form... let this be a lesson that a little research can be a good thing before starting what would otherwise be a years long project. 
20110823, 04:20  #3 
Romulan Interpreter
Jun 2011
Thailand
8,963 Posts 
Thanks all of you. You made my day. I was really upset about losing that amount of work.
Now what's left, is for me to understand what fivemack did there... any links to a good article that explains the math? Should I try second phase with that poly? Or you did that already, too? I see there are no factors reported yet on factordb. [edit] whoops, forgetit! just now I understood what's going on and how stupid I am (about the sieving part). I will give it a try, I don't like to let the things unfinished. 1500 hours... hmmm... Last fiddled with by LaurV on 20110823 at 04:27 
20120221, 03:57  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·4,591 Posts 
SNFS polynomials via lindep
Quote:
For example, Fib(1455); 1455 is divisible by 15, so one could guess that after reducing by Fib(1455/3) and Fib(1455/5) one would expect an octic; and furthermore with a=fib(96); b=fib(98) a symmetric octic, reciprocally turned into a quartic... So let's simply call lindep et voila: Code:
A=24495087715163847304862540972555926557873273177033866366779348623963231442165114660311395897245268799028268424981674090323016224542563692683069015229462064926081 a=fibonacci(96);b=fibonacci(98); u=a^2+b^2;v=a*b; lindep([A,u^4,u^3*v,u^2*v^2,u*v^3,v^4]) [61, 61, 197, 46, 632, 479]~ Running it now... 

20120221, 04:06  #5  
"Ben"
Feb 2007
3352_{10} Posts 
Quote:


20120221, 08:12  #6  
Dec 2008
10110011_{2} Posts 
Quote:


20120221, 13:07  #7  
Nov 2003
2^{2}×5×373 Posts 
Quote:
Yep! F1049 is only C219 with an SNFS sextic; quite doable with even modest resources. Indeed, one could pick off the first 5 to 10 holes in the Lucas and Fibonacci tables. Unfortunately, right now I don't even have "modest" resources. I have a laptop and 3 PCs....... 

20120221, 13:11  #8 
Nov 2003
7460_{10} Posts 

20120221, 13:19  #9  
Nov 2003
2^{2}·5·373 Posts 
Quote:
F1097 is the first hole now. It is perhaps better done by GNFS (C162) [borderline; the size ratio is .70] By SNFS it is C229; still quite doable. The first Lucas hole is at 1018; Very doable by SNFS (C213) 

20120222, 06:55  #10 
Romulan Interpreter
Jun 2011
Thailand
8,963 Posts 
Ye all putting salt on the cut...

20120222, 19:42  #11  
Nov 2003
2^{2}·5·373 Posts 
Quote:
single i7. 

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