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

• To: mathgroup at smc.vnet.net
• Subject: [mg97152] Re: [mg97047] Re: [mg97037] Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
• From: Andrzej Kozlowski <andrzej at akikoz.net>
• Date: Fri, 6 Mar 2009 04:27:35 -0500 (EST)
• References: <200903031056.FAA02950@smc.vnet.net> <200903041209.HAA27014@smc.vnet.net> <47F63136-6342-4418-8AF4-AE3231F9A099@mimuw.edu.pl> <7F2524A9-F9A3-40E8-A7DF-9A44E4BD1350@mimuw.edu.pl>

```Actually I have to change my mind again.

Cohen and Selfridge showed that the twenty-six-digit number,
47,867,742,232,066,880,047,611,079
is prime and neither the sum nor the difference of a power of 2 and a
prime.

Quoted from:

Thus the original conjecture was false (note that the famous result of
Erdos which I mentioned in my first post on this topic deals only with
obstinate number and so does not show this).

Andrzej Kozlowski

On 5 Mar 2009, at 12:26, Andrzej Kozlowski wrote:

>
> 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

```

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