20201107, 00:51  #23  
Feb 2019
China
59 Posts 
Quote:
my code to find F36 factor Code:
Clear["Global`*"]; n=36;(*the nth fermat number*) Do[p=k*2^(n+2)+1;(*probable factor*) If[PrimeQ[p],(*factor must be a prime*) aa=Mod[PowerMod[2,2^n,p]+1,p];(*nth fermat number mod p*) If[aa==0,(*if p is a divisor of fermat number*) Print[{k,p}]; Break[]]], {k,1,100}] Code:
{10,2748779069441} Last fiddled with by bbb120 on 20201107 at 01:11 

20201107, 14:48  #24  
Feb 2017
Nowhere
2^{4}·283 Posts 
Quote:
Code:
? n=1;d=2^38;for(k=1,10000000,n+=d;if(gcd(n,210)==1,r=Mod(2,n);for(i=1,36,r=r^2);if(r==1,print(k" "n)))) 10 2748779069441 3759613 1033434552359452673 For m = 36, we find (results here listed in form odd k, n such that k*2^{n} + 1 divides F_{m}, when, who). 5, 39, 1886, P. Seelhoff 3759613, 38, 02 Jan 1981, G. B. Gostin & P. B. McLaughlin Note well the dates, and the first "who" on that second find. I will pass on seeking the hardware that was used. It should be clear that the most important component is the gray grease in peoples' skulls. Last fiddled with by Dr Sardonicus on 20201107 at 14:54 Reason: Insert missing comma, right paren 

20201107, 15:11  #25  
Mar 2019
9D_{16} Posts 
Quote:


20201107, 18:58  #26 
"Gary"
Aug 2015
Texas
2^{6} Posts 
Phil and I each discovered the F36 factor independently, and then collaborated on the Math Comp paper. I used a Motorola MC6800 processor based system where the hardware, OS, assembler and Fermat search program were all homemade. It was a fun extension of a project I started in grad school.

20201107, 19:21  #27 
Apr 2010
Over the rainbow
11×233 Posts 
Fermat factor are of the form k*2^n+1
You have to either chose a low k, and a large n or the opposite. A very high k and a reasonnable n. either are unlikely. 
20201107, 19:53  #28  
"Robert Gerbicz"
Oct 2005
Hungary
2672_{8} Posts 
Quote:
And if it is prime then check if this divides a Fermat number or not. Last fiddled with by R. Gerbicz on 20201107 at 19:54 

20201107, 20:25  #29  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9417_{10} Posts 
Quote:
By the way,  this reminds my doubts when I heard that there was a theory that the Fermat div probability was not merely ~1/k, and some folks invested a few billiontrillion GHzhours in demonstrating that ...it probably may not be so. But that's probably a good topic for another, separate thread. Quote:


20201107, 22:29  #30 
"Mike"
Aug 2002
2·4,073 Posts 

20201108, 00:27  #31  
Feb 2019
China
59 Posts 
Quote:
my mathematica code can get factor correctly, and it is efficiently ,maybe not the most efficiently. 

20201108, 00:33  #32  
Mar 2019
235_{8} Posts 
Quote:
Of course, you're encouraged to read, understand, and potentially optimize the code for those programs too! 

20201110, 00:20  #33 
Feb 2019
China
59 Posts 
do you find any factor of fermat number ?

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