Even Perfect numbers $n$ with $n+1$ prime The set $S$ of even perfect numbers $n$ such that $n+1$ is a prime number contains
$$
6,28,33550336,137438691328
$$
Latter number found by Joerg Arndt, corresponds to $M_{19}$ (mersenne)
Question:  Is $S$ reduced to these $4$ numbers.
New: Joerg Arndt checked up to exponent $110503$ that the corresponding number $n+1$
is composite. (Improved $19$ to $110503$).
Which function of $x$ migh describe well  the size of the set of elements in $S$ less than $x$
divided
by the size of the set of all even perfect numbers less than $x$; mainly with big $x.$
So, I am asking for relative size not absolute size.  E.g., if I were asking
for  relative density of the prime numbers congruent to $3$ modulo $4$: I do not want to use
the big machinery of the prime number theorem, or Dirichlet's Theorem to deduce how many should be there. I just want (in these case) to know how to describe in terms of $x$
number of primes congruent to $3$ modulo $4$ and less than $x$
divided by
number of primes less than $x$
How many such numbers $n$ we may
expect inside the known 47 perfect numbers ?
 A: The numbers involved are pretty huge - have you tried all the Mersenne primes' perfect numbers yet?  
The other answer might be referring to Wagstaff's conjecture about the number of these primes being less than $e^{\gamma}/\log(2) *\log(\log(x))$; see e.g. here, here, or here for some references (some better than others).
I would imagine that this would be helpful in solving this, but gives a sense of just how hard it would be to prove anything.
A: There's a conjecture (for which I can't find a source now) that the number of Mersenne primes $2^n-1$ with $n < x$ is $c \log x$ for some constant $c$.  Differentiating this, the "probability" that $2^n-1$ is prime is about $c/n$. (This is unconditional; that is, I'm not assuming $n$ is prime.) 
The even perfect numbers are exactly of the form $2^{n-1}(2^n-1)$ with $2^n-1$ prime.
So the "probability" that $2^n-1$ and $2^{n-1} (2^n-1) + 1$ is prime, assuming independence, is $c/n$ times the probability that $2^{n-1} (2^n-1) + 1$ is prime. $2^{n-1} (2^n-1) + 1$ is roughly $2^{2n}$, so by the prime number theorem its "probability" of being prime is about $1/log(2^{2n})$, or again a constant divided by $n$. That is, the "probability" that $2^n-1$ is prime and the corresponding number is one less than a prime is $c/n^2$; since $\sum_{n \ge 1} cn^{-2}$ is finite this leads us to suspect that there are finitely many solutions.
Of course none of this is anywhere near being a proof...
A: I used an awk program to generate congruences on n such that, if the Mersenne exponent n satisfied such a congruence, then the corresponding candidate had a small prime factor, which usually was smaller than the candidate prime.  Using moduli up to 4800, I found that the candidate corresponding to 216091 exponent was a multiple of 4673, most of the remaining candidates were divisible by smaller primes.  At this writing, 61, 1279, 23209, and 20996011, are the exponents whose corresponding candidates may be prime, if I didn't foul up the coding.  So my guess is: at most eight.
Gerhard "Ask Me About System Design" Paseman, 2011.04.25 
A: There is a way to get the answer if instead of using the usual formula for perfect numbers $2^{k-1}(2^k-1)$, we use the following one: $(3n+10)*(3n+11)/2$ with n odd and not necessarily a prime. Then we are basically looking for solutions of $$(9n^2 + 63n + 110)/2 + 1 = 3k + 8$$ or $$(9n^2 + 63n + 110)/2 +1 = 3k+10 $$ with $(3k+8)$ and $(3k+10)$ primes. Note that $(2^k-1)$ can only be a prime of the form $(3k+10)$.
The first equation becomes $$3n^2 + 21n + 32 - 2k = 0$$ The discriminant $d=24k+57$ and we ask that it be a square $m^2$. To get a prime with $(3k+8)$, we need k to be odd. So we will only consider the odd values of k that make d a square. Basically $d=24k + 57=m^2$.
We can check that $k=7$ provides the a value of $n=-1$ that gets us the prime $p=29$. Another pair of values of $(k,n)=(43,2)$ produces a prime $p=137$ but 136 is not a perfect number, just like the pair $(k,n)=(1,-2)$ which produces the prime $p=11$. We cannot use brute force or guessing ( trial and error ) for large perfect numbers. So we need to find a general solution of the diophantine equation $d=24k+57=m^2$.
When fed to the online Dario Alpertron diophantine solver, we get 4 solutions, two of which must be discarded right away because they provide only even values for k. The two other solutions are given by $$m=24t+9$$ and $$k=24t^2 + 18t +1$$ The other solution is given by $$m=24t + 15$$ and $$k=24t^2 + 30t +7$$ Now, we can start a search for solutions  that give both a prime $p=(3k+8)$ and a corresponding value of n that produces a perfect number. This task is better done by writing a program and sadly I can't code.
Keep in mind that the k values that make the discriminant a square do not necessarily make $(3k+8)$ a prime and those that make $(3k+8)$ a prime do not necessarily correspond to a perfect number.  
Note: I cannot comment to answer any question since I have no reputation but my email address is on my profile if needed.
