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

**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>

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

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