20140102, 11:15  #1 
Apr 2010
Over the rainbow
2^{4}×3×7^{2} Posts 
question about a chain of primes
suppose that k*2^x+1, k*2^(2x+1)+1,k*2^(4x+3)+1.... are prime. How long can this chain can be, and are they rare?
24051981*2^24 + 1 is prime 24051981*2^49 + 1 is prime 24051981*2^99 + 1 is prime 24051981*2^199 + 1 isn't prime so that make that one 3 term long. 
20140102, 12:44  #2  
Nov 2003
2·3·17·73 Posts 
Quote:
You are in the wrong forum. And yes, your question can be answered. Read Richard Guy's Book: Unsolved Problems in Number Theory Last fiddled with by R.D. Silverman on 20140102 at 12:51 Reason: Added comment 

20140102, 16:23  #3  
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
5·1,811 Posts 
Quote:


20140102, 16:52  #4 
Romulan Interpreter
Jun 2011
Thailand
5^{2}·7^{3} Posts 
It seems that the "construction" discussed is not a Cunningham chain, as those numbers do not "nearly doubles" every step.
From few pari tests, there are plenty of chains in cause for length 6, they are easy to find for small numbers. Length seven starts with k=969095 and n=1. It seems they can have arbitrary length. Last fiddled with by LaurV on 20140102 at 16:59 
20140102, 17:12  #5  
Nov 2003
16426_{8} Posts 
Quote:
It is trivial to see that such chains of arbitrary but finite length exist if a certain very very well known and very much believed conjecture is true. 

20140102, 17:50  #6  
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
5×1,811 Posts 
Quote:
Anyway, length8 chain appears at p=209556883, and length9 chain appears at p=1991589679 Last fiddled with by Batalov on 20140102 at 18:15 

20140102, 18:23  #7  
Nov 2003
2·3·17·73 Posts 
Quote:
Most of the time when you try to discuss mathematics you say something stupid. These numbers certainly do "nearly double" every step. Or did you fail to take (or pass) first year secondary school algebra? And this sequence most definitely IS a Cunningham chain. (of the second kind) 

20140102, 18:54  #8 
Aug 2006
16E6_{16} Posts 

20140102, 19:11  #9 
Nov 2003
2·3·17·73 Posts 

20140103, 00:21  #10 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
5·1,811 Posts 
Length10 firechain starts with 16115712390779.
8057856195389*2^(2^k1)+1, 1<=k<=10. Code:
16115712390779 64462849563113 1031405593009793 264039831810506753 17304114417533370499073 74320605509547915282165438349313 1370973189237758456882189649193509622387878702088193 466518001818972126590484055297138550142720146550254423311785848574622370435037312216727553 54019094097436859658969629732306068413914220877860574426388607800345512773832031510172556937535694515382891590803380241493774418701828974532859206254646507553970716673 724277638207929266930253555137584980350438559225752130475243057017776657731820280872850742890001087060442445989121458534781350925486567003466887539158918517589176487344301914058802900339316416322436412544887150584702148836067366128236669774778238376700909967366375371470590291086953424789414347049137985049179149421248513 
20140103, 00:55  #11 
Apr 2010
Over the rainbow
2^{4}·3·7^{2} Posts 
Thank you Batalov and Laurv for those example. Now , an important question : Can any of those number Fermat factors?
Last fiddled with by firejuggler on 20140103 at 01:06 
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