Brian Beesley wrote:

> Date: Sun, 23 Nov 2003 08:53:55 +0000
> From: "Brian J. Beesley" <[EMAIL PROTECTED]>
> Subject: Mersenne: Generalized Mersenne Numbers

> Define GM(a,b) = a^b-(a-1), so GM(2,b) = M(b); also GM(a,1) = 1 for all a

What about


For p=2 we have GM(2,n)=2^n-1^2=M(n)

For a prime p>2 we have
GM(p,mp)=p^mp-(p-1)^p= (p^m)^p-(p-1)^p
so it is divisible by p^m-(p-1) (for m=1 it equals 1). I have no idea how
to prove (if it is possible to prove) that GM(p,n) is composite for
composite n. However for p=3 we have (p-1)^p=2^3=8, so n should be greater
than 1 and
GM(3,2)=9-8=  1 (not prime, but not composite, too)
GM(3,3)=27-8=19 prime
GM(3,4)=81-8=73 prime
GM(3,5)=    235= 5*47
GM(3,6)=    721=7*103 (as should be, see above)
GM(3,7)=   2179 prime
GM(3,8)=   6553 prime (for composite exponent!)
GM(3,9) divisible by 3^3-2=25
GM(3,10), GM(3,11) composite 
GM(3,12) divisible by 3^4-2=79
GM(3,13) composite

For p=5 we start from n=5 since 4^5=1024, whereas 5^4=625

GM(5,n) composite for n=5,6,8,9,10 (div. by 25-4=21) 
for n=7 and 11 -- prime 

What happens if p is composite? [for p>3 we always start from n=p since
in this case p^p > (p-1)^p and p^{p-1}<(p-1)^p, p=3 is the exception]

GM(4,n) is composite for n<12
GM(6,11) is prime!

Maybe it is caused by fact that p-1=5 is prime?
What about more general meresennes MGM(p,q,n)=p^n-q^p, p,q prime (or not)?
q^p is related to small Fermat theorem (I don't remember exactly how;
probably q^(p-1) mod p =1 for prime p).

As regards the analog of the LL test, I only remember that Lucas series
is somehow related to Fibonacci series (see D.Knuth and others "Concrete
mathematics"). Is similar series are related with
a^b-(a-1) or a^b-(a-1)^a? 

W Florek 

The most important: CONGRATULATIONS to Micheal, George and Scott and
many, many others.

Wojciech Florek (WsF)
Adam Mickiewicz University, Faculty of Physics
ul. Umultowska 85, 61-614 Poznan, Poland

phone: (++48-61) 8295033 fax: (++48-61) 8295167

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