(2^n+1)/3

[Gottfried Helms]

Am 20.05.05 18:02 schrieb Richard Guy:

> (a)  p = 2^k \pm 1  or  p = 4^k \pm 3
> 
> (b)  M_p  is prime
> 
> (c)  (2^p + 1)/3  is prime.
> 
> [It's not necessary to assume that  p
> is prime.  M_p is the Mersenne number
> 2^p - 1]             R.
> 
given

  (b) M_p   [= 2^p-1]  is prime

 then it follows, that p must be prime.

But anyway, I was not really clear:

> On Fri, 20 May 2005, Gottfried Helms wrote:
> 
>>
>>This is immediately true for a mersenne-prime q, where
>> q = 2^p-1

I should have written

 " let's see, what happens, if q is a mersenne prime
   q = 2^r - 1, too.
 "
where "r" is different from "p" to avoid confusion.

I'm just looking, what happens, if that both strong
restrictions are made, to proceed to weaker(but more
complicated) examples then, and check, whether my
assumptions about these lengthes give a useful base
to discuss that conjecture.


>>
>>Possibly using the 1st condition of the "Bateman..."-conjecture
>>one can narrow this down to a *required* condition by just algebraic
>>manipulation of the above formulas - always on the basis of
>>the -assumed- uniqueness cycle-length of mersenne-primes.
>>
>>I'll try with this a bit today.
>>
>>
>>Gottfried Helms

Gottfried Helms