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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:… – 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

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!

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