Let $(s_n)_{n\in\mathbb N}$ be defined as follows: For $n\in\mathbb N$, $s_n:=2+3+5+\cdots+p_n$ is the sum of the first $n$ prime numbers (e.g.: $s_1=2$, $s_2=5$, $s_3=10$, $s_4=17$, $\ldots$).
Let $\pi_\sigma(n):=\#\{s_k\mid k\leq n\text{ and }s_k\text{ is prime}\}$ (e.g. $\pi_\sigma(1)=1$, $\pi_\sigma(2)=2$, $\pi_\sigma(3)=2$, $\pi_\sigma(4)=3$, $\ldots$).
Comparing $\pi_\sigma(n)$ with $\frac{n}{2\ln(n)}$ ($\ln$ may denote the natural logarithm) using computer power I suspect $$\pi_\sigma(n)\sim\frac{n}{2\ln(n)}.$$
The following picture compares the two functions on the range from 1 to 50.000.000. The blue line is $\pi_\sigma(n)$ and the orange line is $\frac n{2\ln(n)}$.
$\pi_\sigma(n)$ and $\frac n{2\ln(n)}$ on the range from 1 to 50.000.000" />
My question is: Is this true? Has there already been research made on this question? Does anyone perhabs has prooven this conjecture already?
Until now, I just came up with heuristic arguments to deduce this formula (see below). But these arguments assume for example prime numbers to be randomly distributed within the natural numbers in such a way, that their stochastic distribution is given by the prime number theorem.
Let $\varepsilon>0$. Then there exists an $N\in\mathbb N$ and an $\varepsilon'>0$ (converging to $0$ iff $\varepsilon$ converges to $0$), such that for every $n\geq N$ $$n^2\sim\sim\sum_{i=1}^ni<s_n<\sum_{i=1}^ni^{1+\varepsilon'}\sim\sim n^{2+\varepsilon},$$
where $\sim\sim$ may denote asymptotic equivalence up to a constant.
Now, let us argue heuristically, assuming the following:
- The sequence $s_n$ is an increasing random sequence.
- The "probability" of a random number $k\in\mathbb N$ being prime is given by $\frac k{\ln(k)}-\frac{k-1}{\ln(k-1)}$.
We obtain (all up to $\sim\sim$) $$\sum_{i=1}^n\left(\frac{i^2}{\ln(i^2)}-\frac{i^2-1}{\ln(i^2-1)}\right)<\pi_\sigma(n)<\sum_{i=1}^n\left(\frac{i^{2+\varepsilon}}{\ln(i^{2+\varepsilon})}-\frac{i^{2+\varepsilon}-1}{\ln(i^{2+\varepsilon}-1)}\right)$$ and consequently $$\frac12\sum_{i=1}^n\frac{1}{\ln(i)}=\sum_{i=1}^n\frac{1}{\ln(i^2)}<\pi_\sigma(n)<\sum_{i=1}^n\frac{1}{\ln(i^{2+\varepsilon})}=\frac1{2+\varepsilon}\sum_{i=1}^n\frac{1}{\ln(i)}.$$ As $\sum_{i=1}^n\frac{1}{\ln(i)}\sim\int_2^n\frac1{\ln(x)}dx\sim\frac{n}{\ln(n)}$ we conclude $$\frac{n}{2\ln(n)}<\pi_\sigma(n)<\frac{n}{(2+\varepsilon)\ln(n)}.$$
This is a way to deduce the formular heuristically. However, we assumed the sequence $s_n$ to be an increasing random sequence. Why should not $s_{2n}$ be as well an increasing random sequence? This leads to a problem (thank you for the comments @Brian Hopkins, @Joshua Stucky):
Besides $s_1$, all $s_{2n+1}$ are even and therefore not prime. Hence, if we would count for $\pi_\sigma$ only the odd $s_{2n}$ except every $s_n$ (as so done by the 2018 paper indicated by @Brian Hopkins), it would be asymtotic equivalent to $\frac n{\ln(n)}$ instead. This means, the elements of the sequence $s_{2n}$ are "twice as likely" to be prime as we would expect using the prime number theorem. Why should the heuristic argument (2.) work for the increasing random sequence $s_{n}$, but not for the increasing random sequence $s_{2n}$?
