I gave this one to George, he gave a pretty good explanation, but I wanted to 
see what everyone else thought.  I bet He's right, though

>primes of the form (s# + 1), (s! +1 ), (k*(s!) +1) or (k*(s#)+1)  should make
>good exponents, at least in theory.  2^p -1 can be turned into 2^p -2 +1, and
>that turns into 2(2^(p-1)-1) + 1  if p is one of the forms above, then this
>turns into 2(2^s# -1)+1, 2(2^s!-1)+1 , 2(2^k*(s#)-1)+1 or 2(2^k*(s!)-1) +1.
>the 2^s#-1, and related forms must have a lot of factors, like (2^2)-1 ,
>(2^3)-1 , etc.  These are multiplied together, and one is added.  This
>virtually forms the p# + 1, Euclid formula for showing an infinite number of
>primes.  I hope you haven't heard this before.  If you did tell me.  I wonder
>who would have discovered this.  Probably a great prime hunter of the past.

His Response:

I'm not really sure how this helps.  Factors of Mersenne numbers must be
of the form 2kp+1.  Apparently, you are looking for special forms of p
where more of the 2kp+1 values are composite.  An interesting idea,
I'll think about it some.  Remember, I'm more programmer than mathematician.

My Response:
>Every Mersenne prime is of the form (2^n -1).  If n is composite, with
>factors, t and s, the (2^n - 1) is 0 mod (2^t - 1), and (2^s - 1).
>So, if n has a lot of factors, like the primorals or the factorials, it
>should be divisible by a lot of numbers.  If we multiply this by 2 we still
>have a number full of factors.  If we add 1 to this then the resulting number
>is 1 mod all those factors.  So, that new number is relatively prime to all
>those factors, and so has a good chance of being prime.

His response:

Your analysis is good as far as it goes.  However, we already know
that Mersenne numbers have factors of the form 2kp+1.  Thus, the
factors you eliminated may already have been eliminated by the 2kp + 1
criteria.  If that is the case, and I suspect it is, then there is no gain.

Anyone want to comment, or give any idea's?  Just a little something to cheer 
up your day.  I personally think his suspicions could be confirmed, so it 
should be (k*2^n) - 1 to help with this, for those of you who are searching 
for primes other than Merensee's.  
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