20201005, 16:47  #309 
Romulan Interpreter
"name field"
Jun 2011
Thailand
3·7·467 Posts 
Yeah, We need a "Go Ryan! Go!" icon (a runner or something, haha)

20201006, 13:59  #310 
Jun 2012
Boulder, CO
517_{8} Posts 
Thanks! PFGW is still running. Looks like it divides a few other things too.
Code:
7*2^18233956+1 is a Factor of F18233954!!!! (163569.267086 seconds) Special modular reduction using allcomplex AVX512 FFT length 1008K, Pass1=2304, Pass2=448, clm=1, 8 threads on 7*2^18233956+1 7*2^18233956+1 is a Factor of GF(18233952,7)!!!! (174791.088006 seconds) Special modular reduction using allcomplex AVX512 FFT length 1008K, Pass1=2304, Pass2=448, clm=1, 8 threads on 7*2^18233956+1 Special modular reduction using allcomplex AVX512 FFT length 1008K, Pass1=2304, Pass2=448, clm=1, 8 threads on 7*2^18233956+1 7*2^18233956+1 is a Factor of xGF(18233954,7,2)!!!! (334431.512631 seconds) 
20201006, 20:41  #311 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
22610_{8} Posts 
Footnote: Btw, periodically I think (and then I postpone thinking about that) that PFGW could use quite a bit of combinatorial acceleration:
when any single base b test has led to a positive result (e.g. in the case, 2 and 7)  their combination will also lead to hit (but m value will need to be found; there is a cool but predictable structure; b^2^(N+delta) (where delta is increasing from a small negative to zero e.g. {20..0}) can arrive at 1 or directly into 1, and then will remain at 1 during the next squarings  and only very rarely the two bases might poison each other and not produce a hit). Illustration: of course still to come out of this xGF computational run is 7*2^18233956+1 is a Factor of xGF(...,7,4), too. And if another base is hit, e.g. 5, then there will be a constellation of xGF hits that will combine 2 and 5, 2 and 7, 5 and 7. P.S. C.C. has received the email about Fermat divisor and adjusted official remark, and then the system rearranged top lists. Congrats! 
20201006, 22:39  #312 
"Robert Gerbicz"
Oct 2005
Hungary
2×3×11×23 Posts 
As I can remember pfgw is doing this (or something like that, the constant=50 is not that interesting,
we'd catch all known factors if n is not small: n>50): let N=k*2^n+1 and r(u)=u^(2^(n50)) mod N compute it for u=2,3,5,7,11 and store these. Then you can check very easily all xGFN(nd,a,b) mod N=a^(2^(nd))+b^(2^(nd)) mod N for a,b<=12 and d<=50. So basically the extra cost of these xGFN divisibility checks is only four times of a single prp test, because u=2 comes for free. Could we check these times? Last fiddled with by R. Gerbicz on 20201006 at 22:41 Reason: grammar 
20201010, 20:49  #313  
"Jeppe"
Jan 2016
Denmark
5^{2}×7 Posts 
Quote:
Congratulations with this remarkable discovery! Has the above PFGW run completed by now? I see no User comment mentioning the GF(, 7) divisor on https://primes.utm.edu/primes/page.php?id=131289. I do not see your prime on the specific GF(, 7) page http://www.prothsearch.com/GFN07.html. Also, on the page http://www.prothsearch.com/GFNfacs.html, your prime does not appear even when it divides GF(, 7) and xGF(, 7, 2) (and possibly more, I do not know). I really think you should inform the world of your discoveries. On several occasions, I (and others) have had to run the test for ((x)G)F divisibility myself. On the last page I linked, you will see that the credit has been awarded to me (I am "Nielsen" there). In other cases it is S. Batalov or L. Joseph who got the credit. See also your primes on GFN12.html etc. Search for "Propper" on those pages. If you do not "disclose" the information, someone is going to run the PFGW xGF test again, and submit the results, and take the credit. /JeppeSN 

20201011, 03:56  #314  
"Gary Gostin"
Aug 2015
Texas, USA
5×13 Posts 
Quote:


20201011, 12:18  #315 
"Jeppe"
Jan 2016
Denmark
5^{2}×7 Posts 
Very good. Thank you. So in this case the number will be carefully checked and double checked. But I also wrote to Ryan P. to make sure future primes are not forgotten. /JeppeSN

20201012, 07:23  #316  
"Gary Gostin"
Aug 2015
Texas, USA
5·13 Posts 
Quote:
Code:
7 * 2^18233956 + 1 divides GF(18233952,7) 7 * 2^18233956 + 1 divides xGF(18233953,7,4) 7 * 2^18233956 + 1 divides F18233954 7 * 2^18233956 + 1 divides xGF(18233954,7,2) 7 * 2^18233956 + 1 divides xGF(18233954,8,7) 7 * 2^18233956 + 1 divides xGF(18233954,11,9) 7 * 2^18233956 + 1 divides xGF(18233955,9,5) 7 * 2^18233956 + 1 divides xGF(18233955,10,9) 7 * 2^18233956 + 1 divides xGF(18233955,11,5) 7 * 2^18233956 + 1 divides xGF(18233955,11,10) 

20201012, 20:10  #317  
"Jeppe"
Jan 2016
Denmark
5^{2}×7 Posts 
Quote:


20201012, 21:34  #318  
Feb 2017
Nowhere
13F2_{16} Posts 
Quote:
N  GF(18233952,7) we obtain N  7^(2^18233953)  1. Since also N  F18233954, we obtain N  xGF(18233953,7,4). From N  7^(2^18233953)  1 we obtain N  7^(2^18233954)  1. Again applying N  F18233954, we obtain N  xGF(18233954,7,2) Since N  F18233954, F18233954  8^18233954 + 1, and N  7^(2^18233954)  1, we obtain N  xGF(18233954,8,7). I haven't figured out an easy path to the other ones. 

20201012, 22:20  #319  
"Jeppe"
Jan 2016
Denmark
5^{2}·7 Posts 
Quote:
Let p = k * 2^n + 1 be a prime that divides a Fermat number. From the formula, we can write k = 1/2^n (mod p). But the order of 2 (mod p) is a power of two, and clearly the order of 1 is as well, so this shows that the order of k (mod p) is a power of two. So p divides a GF(, k). So we could tell in advance, once we knew 7*2^18233956+1 divided an F(), that it also divided a GF(, 7). /JeppeSN 

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