20200123, 16:54  #287 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·3·797 Posts 
We would be amiss to not notice this new find that has just shown on radars...
13 · 2^{5523860} + 1 divides Fermat F(5523858) Congratulations to Scott and the numerous PrimeGrid miners! 
20200123, 17:39  #288  
Sep 2002
Database er0rr
3,863 Posts 
Quote:
Last fiddled with by paulunderwood on 20200123 at 17:43 

20200124, 02:41  #289 
Romulan Interpreter
Jun 2011
Thailand
10011000110000_{2} Posts 

20200124, 13:52  #290 
"Jeppe"
Jan 2016
Denmark
A8_{16} Posts 
Good old F(5523858): I always thought it would be composite, but I did not expect such a tiny prime divisor! /JeppeSN

20200124, 15:07  #291 
Feb 2017
Nowhere
2^{3}·7·89 Posts 
Once upon a time, long long ago, I noted that if n > 2 and k < 2^{n+2} + 2, the fact that N = k*2^{n+2} + 1 divides F_{n} in and of itself proves N is prime. In the case at hand (n = 5523858), k = 13 satisfies this condition.
Last fiddled with by Dr Sardonicus on 20200124 at 15:08 Reason: Omit unnecessary words! 
20200124, 20:35  #292 
Banned
"Luigi"
Aug 2002
Team Italia
4827_{10} Posts 
Any more infos about the discovery? like the surname of Scott, or the setup he used for the search, or the time he devoted to it?
Luigi  P.S. Dr. James Scott Brown. Last fiddled with by ET_ on 20200124 at 20:47 
20200124, 20:45  #293  
Mar 2019
2^{3}·5^{2} Posts 
Quote:


20200124, 22:40  #294  
Einyen
Dec 2003
Denmark
6147_{8} Posts 
Quote:
http://primegrid.com/forum_thread.php?id=8778 https://www.primegrid.com/workunit.php?wuid=638323572 https://www.primegrid.com/show_user.php?userid=1178 

20200124, 23:41  #295 
"Jeppe"
Jan 2016
Denmark
250_{8} Posts 
It is the same Scott Brown who found another Fermat factor, 9*2^2543551+1, back in 2011, in a similar way.
In PrimeGrid, the participants detect the primality (two persons do it concurrently, the one finishing first is declared the finder). Whether the new Proth prime divides any Fermat number (and/or generalized Fermat numbers with bases at most 12) is detected by PrimeGrid's server, not the participant's computer. The link posted by ATH shows the timing (primality was reported "22 Jan 2020  15:00:58 UTC") and some hardware ("Intel(R) Core(TM) i74790 CPU @ 3.60GHz [Family 6 Model 60 Stepping 3]"). The other participant was Stefan Larsson (returned "22 Jan 2020  15:11:39 UTC"). At some point PrimeGrid will publish an official announcement (PDF). This was PrimeGrid's first Fermat divisor in five years. They recently introduced a new subproject that focuses on Proth primes with low "k" (the odd multiplier) because that gives higher probability of Fermat divisors. This approach was recommended by Ravi Fernando and others. /JeppeSN 
20200125, 14:28  #296 
Feb 2017
Nowhere
2^{3}·7·89 Posts 
Just for fun, I looked for small prime factors of the numbers k*2^5523860 + 1, k = 1 to 12. For all but three of them, the smallest factor can be found mentally. For two of the remaining three, the smallest factor is still quite small.
k = 1 p = 17 k = 2 p = 3 k = 3 p = 14270779 k = 4 p = 5 k = 5 p = 3 k = 7 p = 2625617 k = 8 p = 3 k = 9 p = 5 k = 10 p = 11 k = 11 p = 3 k = 12 p = 7 For the remaining value, 6*2^5523860 + 1, I didn't look far enough to find any factors, but I didn't look all that far. Has (as I suspect) someone already found a factor by trial division, or otherwise shown it to be composite? 
20200125, 17:04  #297 
"Jeppe"
Jan 2016
Denmark
10101000_{2} Posts 
PrimeGrid's official announcement is here and can be found in this thread of theirs. /JeppeSN

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