# 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."

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Nice but probably very hard problem! If the conjecture is true, that roughly means that the very "special" numbers $R_k$ are about as likely to be prime as any odd number. Sounds quite plausible, and sadly beyond current techniques. – André Nicolas Aug 10 '11 at 2:25
Cross post to MO: mathoverflow.net/questions/72536/… – Grumpy Parsnip Aug 10 '11 at 2:30
I wouldn't call $R_n$ a recursion since its definition never refers back to itself. It's just a sequence derived from the sequence of primes. – robjohn Aug 10 '11 at 2:36
Oh - I was about to post a heuristic too, but I just read MO and realized that it's already been done. I feel so slow. – mixedmath Aug 10 '11 at 3:00
I've taken the liberty of providing a more informative title, and changing "recursion" to "sequence". Also, in view of the problem being posted to MO, I vote to close it here. – Gerry Myerson Aug 10 '11 at 6:13