20201112, 06:00  #34 
Apr 2010
Over the rainbow
2·5·11·23 Posts 
Any factor of this would be the largest prime ever found.

20201112, 07:00  #35 
"Curtis"
Feb 2005
Riverside, CA
11106_{8} Posts 
Why is that? Factors would have form 2kp+1, where p is 2^1271, right?
Last fiddled with by VBCurtis on 20201112 at 07:00 Reason: k > p 
20201112, 07:17  #36  
Romulan Interpreter
Jun 2011
Thailand
10010000111011_{2} Posts 
Quote:
Last fiddled with by LaurV on 20201112 at 07:22 

20201112, 07:48  #37 
Apr 2010
Over the rainbow
4742_{8} Posts 
My bad I made a mistake.

20201112, 10:50  #38 
"Viliam Furík"
Jul 2018
Martin, Slovakia
2^{2}×3^{2}×11 Posts 

20201112, 12:29  #39 
"Rashid Naimi"
Oct 2015
Remote to Here/There
7C0_{16} Posts 
Last fiddled with by a1call on 20201112 at 12:46 
20201112, 13:07  #40  
"Viliam Furík"
Jul 2018
Martin, Slovakia
2^{2}·3^{2}·11 Posts 
Quote:
Calculus is calculus, other mathematics is other mathematics. 

20201112, 13:14  #41 
Jun 2003
4,861 Posts 

20201112, 14:19  #42  
Feb 2017
Nowhere
2^{3}×3×181 Posts 
Quote:
1) The smallest prime exceeding F_{33}; in PariGP notation, this could be expressed nextprime(2^(2^33)33 + 2) although PariGP only guarantees a value for which ispseudoprime() returns 1. Not a problem here. Bertrand's Postulate (which has of course been proven) guarantees the existence of a prime between F_{33} and 2*F_{33}. 2) The smallest prime factor of F_{2^(2^34)  2}. 

20201112, 14:57  #43 
Mar 2019
2^{4}×3^{2} Posts 
We can at least write F_{33} as 2^8589934592+1, yes? That may at least help to illustrate just how much larger it is than the largestknown prime, 2^825899331.

20201112, 17:25  #44  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2×4,691 Posts 
Quote:
"binary splitting" algorithm or "binary splitting" algorithm explained Last fiddled with by Uncwilly on 20201112 at 17:26 
