[Math] Asymptotics for primality of sum of three consecutive primes

analytic-number-theorynt.number-theoryprime numbersprime-number-theorem

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 $R_n$ 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, 371, 395, 407, 425, 439, 457, …

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."

Best Answer

I believe proving (or disproving) such a statement is beyond current technology. On the other hand, by a crude heuristics the conjecture must be right: $R_1,\dots,R_n$ are $n$ odd numbers up to $\sim 3n\ln(n)$ which are rather evenly distributed in size and in residue classes. In this range the density of primes among odd numbers is $\sim 2/\ln(n)$, so $R(n)$ should be $\sim 2n/\ln(n)$.

EDIT: As Noam Elkies points out in a comment below, there should be a fine tuning similar to the twin prime constant.