# Sums of primes that are themselves prime

I'm not a math expert so this may be a trivial question; if $p_i$ is the $i$-th prime, let:

$$S(n) = \sum_{i=1}^n p_i$$

be the sum of the first $n$ primes and

$$P(n) = | \{1 \leq i \leq n \mid S(i) \mbox{ is prime} \} |$$

be the number of the sums $S(i), 1 \leq i\ \leq n$ that are prime. Do we have

$$\lim_{n \to \infty} \frac{P(n)}{n} = 0\;?$$

Where can I find a proof?

EDIT: I generated the sequence $P(n)$ and found it on OEIS: Numbers n such that n is prime and is equal to the sum of the first k primes for some k., but there is not a lot of information there.

• See also oeis.org/A013916 although there is not much information there, either. – Gerry Myerson Jan 5 '14 at 16:17
• A first step would be to understand how many of the $S(n)$ are multiples of a given number $d$. Already for $d=3$ I don't know how to determine the number of $S(n)$'s that are multiples of $3$. Such questions seem too hard to answer (although one can of course formulate conjectures). – Lucia Jan 5 '14 at 16:37
• @TerryTao: The sums $S(n)$ alternate in parity. So $P(n)/n$ is trivially at most $1/2$. But what you say would apply to maybe get strictly below $1/2$. – Lucia Jan 5 '14 at 18:39
• Yes, good point. If one uses Maynard (and Banks et al.) to ensure at least six primes amongst the $n+h_i$, then we can get S(i) divisible by 3 for infinitely many even i, and so one should indeed be able to get $P(n)/n$ below 1/2 by a modification of this argument (one has to allow the diameter of the admissible tuple to be as large as a small multiple of $\log n$). – Terry Tao Jan 5 '14 at 18:41
• @TerryTao: I don't understand how one gets a constant below $1/2$ though. My understanding is that the work of Maynard et al does not yet give positive proportion of tuples. I.e. how does one get a positive proportion of values for which $S(n)$ is a multiple of $3$? (Infinitely many is fine, and really goes back to Daniel Shiu.) – Lucia Jan 5 '14 at 18:45

Instead of adding all prime numbers, let us "flip a coin" and include the next prime number with a fixed probability $0<q<1.$
$$S_{\xi}(n)=\sum_{i=1}^n \xi_i p_i,$$
where $p_i$ is the $i$th prime and $\{\xi_i\}$ is a sequence of i.i.d. Bernoulli random variables? The asymptotic needs to be a.s. with respect to the Bernoulli measure.