[Date Index]
[Thread Index]
[Author Index]
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:
http://books.google.com/books?id=1MTcYrbTdsUC&pg=PA176&lpg=PA176&dq=obstinate+numbers+Erdos&source=bl&ots=9E60IbVmZ1&sig=8HdQDoHKBGV4P3WhcjhmK6OjJlQ&hl=en&ei=BtewScTtBNKX_gawmJDJBA&sa=X&oi=book_result&resnum=1&ct=result
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
Prev by Date:
**Re: Re: GraphComplement doesn't work in 7.0**
Next by Date:
**Re: Re: "Do What I Mean" - a suggestion for**
Previous by thread:
**Re: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p**
Next by thread:
**Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p**
| |