20081209, 23:00  #1 
Apprentice Crank
Mar 2006
11·41 Posts 
Does NewPGen have a bug?
I was sieving k=173 from n=1M to n=2M, and I found this message by NewPGen:
p=2215115304221 divides n=1798220. However 2,215,115,304,221 is composite (that number is 8627 * 256765423), so that message should not have popped up. Why didn't NewPGen remove n=1798220 when it was at p=8627 or at p=256765423? In case anyone's wondering, I was using NewPGen version 2.82 on a Pentium 4 computer. The problem also appears on another Pentium 4 machine, so I'm pretty sure it's not a hardware problem. Last fiddled with by MooooMoo on 20081209 at 23:05 Reason: typo :( 
20081209, 23:15  #2  
"Robert Gerbicz"
Oct 2005
Hungary
2^{3}×3×59 Posts 
Quote:


20081209, 23:23  #3 
Apprentice Crank
Mar 2006
703_{8} Posts 

20081209, 23:38  #4 
Apprentice Crank
Mar 2006
11×41 Posts 
Here's the test file that I used, you can try it for yourself to see that p=2215115304211 really does divide n=1798220:
Code:
2215115300000:M:0:2:258 173 1798016 173 1798038 173 1798060 173 1798064 173 1798068 173 1798080 173 1798096 173 1798098 173 1798130 173 1798134 173 1798172 173 1798196 173 1798212 173 1798220 173 1798236 173 1798254 173 1798264 173 1798274 173 1798320 173 1798344 173 1798348 173 1798350 173 1798382 173 1798386 173 1798400 173 1798420 173 1798428 173 1798436 173 1798448 173 1798464 173 1798528 173 1798536 173 1798542 173 1798562 173 1798576 173 1798614 173 1798628 173 1798652 173 1798656 173 1798668 173 1798684 173 1798696 173 1798704 173 1798718 173 1798736 173 1798742 173 1798768 173 1798782 173 1798812 173 1798854 173 1798898 173 1798920 173 1798924 173 1798956 
20081210, 00:48  #5 
Nov 2003
E26_{16} Posts 
In your post #1 you mentioned p1=2215115304221.
In your post #3 you mentioned p2=2215115304211. Observe that p1 != p2. PARI says: Code:
gp > factorint(p1) [8627 1] [256765423 1] // p1 is composite gp > factorint(p2) [2215115304211 1] // p2 is prime (09:46) gp > p1p2 %15 = 10 // p1 != p2 gp> c=173*2^17982201 gp > c%p1 %16 = 1232127380929 // p1 is not a factor of c gp > c%p2 %17 = 0 // but p2 is! Last fiddled with by Kosmaj on 20081210 at 00:54 
20081210, 03:13  #6  
"Mark"
Apr 2003
Between here and the
2·3,001 Posts 
Quote:


20081210, 03:58  #7 
Apprentice Crank
Mar 2006
11·41 Posts 

20081210, 04:54  #8 
A Sunny Moo
Aug 2007
USA (GMT5)
14141_{8} Posts 

20081210, 07:04  #9 
May 2007
Kansas; USA
10239_{10} Posts 
Actually, sr1sieve would be the fastest for one OR two k's. For two k's, just run two instances of sr1sieve, which will eat less CPU time than one instance of sr2sieve.

20081210, 19:13  #10 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
1011001101111_{2} Posts 
what is the crossover point is it 3 ks or more or does it depend on the size of the k or something else

20081210, 19:47  #11  
"Robert Gerbicz"
Oct 2005
Hungary
588_{16} Posts 
Quote:
Code:
b=5684341886080801486968994140701 n=2481892977737648921240994084183 p=3433683820292512484657849089373 

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