Asymptotics for prime sums of three consecutive primes We consider the sequence $R_n=p_n+p_{n+1}+p_{n+2}$, where $\{p_i\}$ is the prime number sequence, with $p_0=2$, $p_1=3$, $p_2=5$, etc..
The first few values of $R_n$ for $n=0,1,2,\dots $ are:
$10, 15, 23, 31, 41, 49, 59, 71, 83, 97, 109, 121, 131, 143, 159, 173, 187, 199, $ 
$211,223,235,251,269,287,301,311,319,329,349,271,395,407,425,439, 457$
$\dots \dots \dots$
Now, we define $R(n)$ to be the number of prime numbers in the set $\{R_0, R_1 , \dots , R_n\}$.

What I have found (without justification) is that $R(n) \approx \frac{2n}{\ln (n)}$
My lack of programming skills, however, prevents me from checking further numerical examples. I was wondering if anyone here had any ideas as to how to prove this assertion.
As a parting statement, I bring up a quote from Gauss, which I feel describes many conjectures regarding prime numbers:
"I confess that Fermat's Theorem as an isolated proposition has very little interest for me, because I could easily lay down a multitude of such propositions, which one could neither prove nor dispose of."

What this says (to me) is that there are twice as many primes in R(n) as in the natural numbers. But, since each R(n) is the sum of three primes, all of which are odd (except for R(1)), then each R(n) is odd. This eliminates half of the possible values (the evens), all of which are, of course, composite.
So, if the R(n) values are as random as the primes (which, as Kac famously said, "play a game of chance"), then they should be twice as likely as the primes to be primes since they can never be even.
As to proving this, haa!
