Conjecture: the sequence of sums of all consecutive primes contains an infinite number of primes Starting from 2, the sequence of sums of all consecutive primes is:
$$\begin{array}{lcl}2 &=& 2\\
    2+3 &=& 5 \\
    2+3+5 &=& 10 \\
    2+3+5+7 &=& 17 \\
    2+3+5+7+11 &=& 28 \\
    &\vdots&
\end{array}
$$
If the $n^\text{th}$ prime is $P_n$, then we can write $S_n=\sum_{i=1}^n P_n$.
I conjecture that the sequence $S_n$ contains an infinite number of primes.
I doubt I'm the first person ever to think this, but I cannot find reference to the idea, nor can I conceive a proof, nor a disproof.  Computationally, it can be verified that
    $$S_{13,932}=998,658,581=P_{50,783,012},$$
and the sequence shows no sign of slowing down.
Can you prove the sequence $S_n$ contains an infinite number of primes?
 A: The recent paper by Romeo Meštrović on arXiv at Curious conjectures on the distribution of primes among the sums of the first $2n$ primes states your conjecture, along with certain details in section $3$ such as an estimate of the number of primes (e.g., that your sequence $S_n$ asymptotically has $\frac{n}{2\log n}$ primes). This paper's other sections examine the distribution of primes in the sequence (using heuristic, computational and analytic arguments) to show they are quite similar to the distribution of primes among the positive integers.
A: As Will Jagy points out, a proof is probably out of reach, but the statement is probably true.
One can give a heuristic argument. The $n$-th prime is roughtly $n\log n$, so $S_n$ is roughly
$$S_n \approx \sum_{i=1}^n i \log i \approx \frac{n^2(2\log n - 1)}{4}.$$
The density of primes around $x$ is approximately $1/\log x$, so the probability that $S_n$ is prime is roughly
$$1/\log\left(\frac{n^2(2\log n - 1)}{4}\right) \approx \frac{1}{2\log n + \log(2\log n-1) - \log 4} \approx \frac{1}{2\log n}.$$
So the expected number of primes among $S_1, \dots, S_n$ is roughly
$$\sum_{i=1}^n \frac{1}{2\log i} \approx \frac{\text{li}(n)}{2},$$
which goes to $\infty$ as $n\to \infty$.
(I just saw that Greg Martin came to the same conclusion in the comments.)
A: Looking for "prime partitions" might help to find prior information for this conjecture.
Another conjecture would be the sum of the gaps of consecutive primes, $g_n = p_{n+1} - p_n$: $$s = \sum_{i=1}^{n}g_i,$$ having an infinite number of primes. Considering the fact that the next prime is $p_{n+1} = s + 2$ one might think it be easy, but it is not.
