Mersenne numbers and the (divisors) property.

[Annette.Warlich at t-online.de]

Well, I think I should add an explanation, why I
said the following:

Am 28.04.05 08:24 schrieb Annette.Warlich at t-online.de:

>  If there would be a general definition for L in a similar
>  form as (1), (which were independent of the mersenne-factorization),
>  or (which would be equivalent) for your above sequence,
>  and this could indeed be expressed with an exponential form in b,
>  then one could use that form to prove the catalan-conjecture
>  in a very simple way - at least concerning powers of primes.
> 
Here is the most simple example, with p=3
It seems to be true that

(0)  L(2^a-1,2) = a

If you could show that
(0.1)   L(p^b,2 ) = L(p,2)* p^(b-1)   // with exceptions for some p

holds for p=3 and N = p^b = 3^b, so that

(1)  L(3^b,2) = 2 * N/3    // assumption

then, if any power a of 2 neighbours 3^b , then

(2.1)    2^a - 1   =  3^b      = N
(2.2)    2^a + 1   =  3^b      = N


For instance let's follow (2.1) :

(2.1)    2^a - 1   =  3^b      = N

then the lengthes of lhs and rhs also must be equal:

(3.1)    L(2^a - 1,2)   =  L(3^b,2)      = L(N)

Plugging (0) and (1) into (3.1):

(3.2)       a           = 2*3^b/b        = L(N)

plugging a from (3.2) into (2.1) :

(3.3)    2^(2*3^(b-1)) -1   = 3^b
(3.4)    4^3^(b-1)     -1   = 3^b

This has the only solution with b=1
 From that follows, that a = 2
Then
        2^2 - 1 = 3^1
which is the only solution.

Analoguously this can be done starting at (2.2).

----------

If one could find a better expression for (0.1) which

- is independent of (0),
- keeps the property of being exponential in b, and
- is valid for all primes p,

then the same way of arguing could be used for any p.

For composite n things are even more complicated, but if
an analoguos formula to that of my previous posting could
be found for composite n, then this would give access to
the catalan conjecture in a very simple way.

Gottfried Helms