# 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?

• very hard to believe a proof is possible, when we can not prove that there are infinitely many primes $n^2 +1$ Jan 13 '14 at 2:55
• I am surprised that the sequence 2,5,17,29,59,101 does not appear in the OEIS. I would recommend adding it. Jan 13 '14 at 3:14
• Standard heuristics would predict that the number of $n\le x$ for which $S_n$ is prime is asymptotic to $x/(2\log x)$ (and computation seems to bear this out). But I agree it's hopeless to expect a proof that there are infinitely many prime $S_n$. Jan 13 '14 at 3:18
• oeis.org/A013918
– user940
Jan 13 '14 at 3:30
• Related: mathoverflow.net/q/153656/12357 Jan 13 '14 at 3:37

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.)

• The bound for $\sum_{p \leq n} p$ you gave is too rough, why not improve it by successive partial sums? Also, perhaps improve the probabilistic arguments by coming up with bounds instead of asymptotics? In any case, the above heuristic can be definitely improved. Jan 13 '14 at 8:50
• @BalarkaSen No doubt it can be improved. Jan 13 '14 at 13:56

The recent paper "Curious conjectures on the distribution of primes among the sums of the first $$2n$$ primes" by Romeo Meštrović at https://arxiv.org/abs/1804.04198 states your conjecture, along with certain details like the estimate of the number of primes, in section 3. This paper's other sections examine the distribution of primes, using heuristic, computational and analytic arguments that they are similar to the distribution of primes among the positive integers.

• Nice reference! Dec 17 '18 at 16:18

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.

• The sum you mentioned is $p_{n+1}-2$, the conjecture you mentioned is therefore equivalent to the twin prime conjecture. Sep 2 '15 at 10:15