Is the sum of the first $n$ primes a prime infinitely many times?

Define the sequence $P(n)=\sum_{i=1}^{n}p_i$, where $p_i$ is the $i$-th prime number.

I observed for some small $n$ that sometimes this sum evaluates to a prime number, for example $P(2)=2+3=5$ and $P(4)=2+3+5+7=17$ and $P(6)=2+3+5+7+11+13=41$.

So it is natural to ask:

Is it true that there is a sequence of natural numbers $\{n_i:i \in \mathbb N\}$ such that all numbers in the set $\{P(n_i):i \in \mathbb N\}$ are prime numbers?

• I don't think we'll be able to answer this – Jorge Fernández Hidalgo Jun 3 '16 at 17:08
• Maybe someone can evaluate this sum for the first, let´s say 1000 (or more) even numbers to see is this sum often a prime number? – Farewell Jun 3 '16 at 17:12
• yeah, thats easy, I'll do it. – Jorge Fernández Hidalgo Jun 3 '16 at 17:12
• up to $n=1$ there are $1$ prime values, this is $100.000000$ percent up to $n=10$ there are $4$ prime values, this is $40.000000$ percent up to $n=100$ there are $10$ prime values, this is $10.000000$ percent up to $n=1000$ there are $76$ prime values, this is $7.600000$ percent – Jorge Fernández Hidalgo Jun 3 '16 at 17:32
• This is a duplicate of (found this now): math.stackexchange.com/questions/636479/… – Farewell Jun 3 '16 at 19:26

Almost certainly yes, but I doubt very much that this can be proven at the current state of the art.

The first few primes that arise are

$$2, 5, 17, 41, 197, 281, 7699, 8893, 22039, 24133, 25237, 28697, 32353, 37561, 38921, 43201, 44683, 55837, 61027, 66463, 70241, 86453$$

The sum of the first $n$ primes is on the order of $n^2 \log n$, and heuristically a number of this size has probability on the order of $1/\log n$ of being prime. Since $\sum_n 1/\log n = \infty$, there ought to be infinitely many. But that's not a proof.

• This enormous "jump" from $281$ to $7699$ is interesting. – Farewell Jun 3 '16 at 17:16
• I wonder if one can define some sort of (independence/equidistribution) conjecture on the prime numbers implying that all those kind of sequences have infinitely many primes (infinitely many twin primes, infinitely many Mersenne primes, etc.) and if the fact that all those sequences have infinitely many primes would imply the Riemann hypothesis – reuns Jun 3 '16 at 21:17

This is not an answer, but it intend to show that there seems to be an infinite number of primes of the form $\sum_{k=1}^n p_k$.

The x-axis in the diagram is $\,^2\!\log n$ and the y-axis is $\,^2\!\log N_n$, where $N_n$ is the number of primes of the form $\sum_{k=1}^m p_k$ where $m\le n$.

up to $n=1$ there are $1$ prime values, this is $100.000000$ percent

up to $n=10$ there are $4$ prime values, this is $40.000000$ percent

up to $n=100$ there are $10$ prime values, this is $10.000000$ percent

up to $n=1000$ there are $76$ prime values, this is $7.600000$ percent

• Should this be a comment? – Pedro Tamaroff Jun 3 '16 at 17:31
• I don't know, it looks better as an answer. – Jorge Fernández Hidalgo Jun 3 '16 at 17:33
• I don't think this actually addresses the question, which is about when the sum of the first $n$ primes is a prime. – Noah Schweber Jun 3 '16 at 20:01
• Well, the OP asked for computational results, I was just trying to help. – Jorge Fernández Hidalgo Jun 3 '16 at 20:12

Not an answer, for the obvious reasons, stated by other people. But here is an attempt I am trying, unsuccessfully so far. There is this book "Problems in Real Analysis. Advanced Calculus on the Real Axis" by:

Now, combining this with Legendre's conjecture, we obtain: $$\left | \sqrt{P(n)} - \sqrt{q} \right | < 1$$ where $q$ is some prime number within the same consecutive perfect squares as $P(n)$.
Obviously, not every $n$ will yield a possible prime to consider, because $P(n)$ is even for odd $n$. But, if by some magic density or Pigeonhole principle argument we could prove that $\left | \sqrt{P(n)} - \sqrt{q} \right |$ is close enough so that $\left | P(n) - q \right | < 1$, that would make it a nice proof, based on Legendre's conjecture of course.