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Re: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p

• To: mathgroup at smc.vnet.net
• Subject: [mg97156] Re: [mg97047] Re: [mg97037] Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 6 Mar 2009 04:28:18 -0500 (EST)
• References: <200903031056.FAA02950@smc.vnet.net> <200903041209.HAA27014@smc.vnet.net> <47F63136-6342-4418-8AF4-AE3231F9A099@mimuw.edu.pl>

On 4 Mar 2009, at 22:40, Andrzej Kozlowski wrote:

> On the other hand, the OP question was a bit different as he  
> included the possibility of "negative prime", in other words his  
> question was: is it true that for every obstinate number x one can  
> find an integer n such that 2^n-x is prime. Actually, one could ask  
> an even stronger question: is it true that for every odd number a  
> there exists an integer n such that 2^n - a is prime? It seems very  
> likely that the answer is yes, and probably the proof is easy but I  
> can't spend any more time on this...


Actually, on reflection I have changed my mind. I don't think it will  
be easy to prove that for every odd a there is an n such that 2^n-a is  
prime. When I wrote that I expected that for every odd a there would  
be infinitely many n such that 2^n -a is prime, but, of course, this  
is not even known for a=1. In view of that I now tend to think the  
conjecture is probably true but very hard to prove.

Andrzej Kozlowski

• References:
  • Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
    • From: Tangerine Luo <tangerine.luo@gmail.com>
  • Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
    • From: DrMajorBob <btreat1@austin.rr.com>