20181221, 07:30  #276  
"Gary Gostin"
Aug 2015
Texas, USA
2×3×11 Posts 
Quote:
Last fiddled with by ET_ on 20181221 at 08:42 

20190129, 10:31  #277 
Banned
"Luigi"
Aug 2002
Team Italia
4834_{10} Posts 
January 28th, 2019
New Fermat factor from FermatSearch! 1527888802614951 . 2^{120}+1 is a Factor of F_{118}!!! Peter Strasser discovered the first Fermat factor of this year! He used George Woltman's mmff program running on his home computer Congratulations to Peter from FermatSearch, for his third factor! 
20190129, 11:32  #278  
"Roman"
Dec 2016
Everywhere
26_{10} Posts 
Quote:


20190326, 04:15  #279 
"Gary Gostin"
Aug 2015
Texas, USA
2×3×11 Posts 
A new factor from FermatSearch
I would like to report the following new Fermat factor:
332,436,749 * 2^9865 + 1 divides F9863 This was discovered March 23 using pmfs on a HPE Superdome X system. In each year from 2013 to 2018 no more than 7 new factors were found. So far in 2019 we are on the same pace: 2 factors in ~3 months. But who knows, maybe we will all get lucky and find many new factors. Might we match the 16 new factors discovered in 2012? 
20190329, 12:53  #280  
Mar 2019
205_{10} Posts 
Quote:
Curious which range(s) you have fully searched? 

20190331, 15:59  #281 
"Gary Gostin"
Aug 2015
Texas, USA
2·3·11 Posts 
Thanks mathwiz!
That depends on what you mean by "fully". For every range assigned to me in the fermatsearch.org Merge Tables, I have tested each Proth number (k * 2^n + 1) that survives the sieve to see if it divides any classic Fermat number. I have NOT tested to see if it divides any GFN or xGFN, since (as I understand it) that usually involves first testing to see if the Proth number is prime or PRP, which my program does not do. 
20190331, 17:23  #282  
"Robert Gerbicz"
Oct 2005
Hungary
2^{4}×5×19 Posts 
Quote:
what you are doing is: let d=k*2^n+1 what survived the sieving process: you calculate a(i)=2^(2^i) mod d, for say up to i<=n+30. (nobody saves the whole "a" array, only the last term). for a cheap prp test you need only: res=2^(d1)=a(n)^k mod d, ofcourse for a prp number you need res=1. For n~10000, k~1e30 the slowdown of the extra prp test is less than 0.5%, in general it is ~T*log(k)/n if the total running time was T. 

20190402, 03:11  #283  
"Gary Gostin"
Aug 2015
Texas, USA
2·3·11 Posts 
Quote:
Itâ€™s difficult to find a complete definition of all the GFN terminology on the web, but several researchers seem to be using the following: Generalized Fermat Numbers (GFN or GF): GF(n,a) = a^(2^n) + 1 Extended Generalized Fermat Numbers (xGFN or xGF): xGF(n,a,b) = a^(2^n) + b^(2^n) Someone please correct me if this is inaccurate. 

20190402, 09:48  #284  
"Robert Gerbicz"
Oct 2005
Hungary
2^{4}·5·19 Posts 
Quote:
Discovered later, but the real slowdown would be higher in your range, because you need also a^(2^i) mod a for a<=12 and i<n, if you want the vintage range for GFN,xGFN. If you are not obsessed with only Fermat factors, then the slowdown would be roughly 4 % with n~10000, sieving depth~50. Don't see here a shortcut, only the "known" gain, let r(a)=a^(2^(n30)) mod d, for prp we already know r(2), then compute r(3),r(5),r(7),r(11) (so where the base is prime) and then with only one mulmod you can get the rest: r(1)=1 (ofcourse) r(4)=r(2)*r(2) mod d r(6)=r(2)*r(3) mod d r(8)=r(2)*r(4) mod d r(9)=r(3)*r(3) mod d r(10)=r(2)*r(5) mod d r(12)=r(2)*r(6) mod d [you need to do this in order, because you also reuse composite index, say r(4) for r(8)] after that you know r(a) mod d for all a<=12. Then for up to m=n1 with only extra 12 (actually 11) mulmods per exponent, and in total 29*11 mulmods you can get a^(2^i) mod d for a=1..12,i=n30..n1. And at each i=(n30)..(n1) check to see if a^(2^i)+b(2^i) mod d = 0 for xGFN (basically checking only GFN gives no speedup). ps. so you have to stop at i=n1, and not at i=n2 (like for Fermat number divisors). 

20191211, 17:47  #285 
Banned
"Luigi"
Aug 2002
Team Italia
12E2_{16} Posts 
New Fermat factor from Gary Gostin!
News from Gary Gostin!
I would like to report a new Fermat factor: ***** 15,249,465,809 * 2^2591 + 1 divides F2587 This was discovered running pmfs on an HPE Superdome X system. Mankind has now discovered 350 factors of Fermat numbers! We should have a party! 
20191212, 09:34  #286 
Romulan Interpreter
"name field"
Jun 2011
Thailand
23124_{8} Posts 
Yeeee! Congrats!

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