I made some numerical simulations about the number of primes which are the sum of 3 consecutive primes ([OEIS A034962](https://oeis.org/A034962)), that is for instance:
$$5+7+11=23$$
$$7+11+13=31$$
$$11+13+17=41$$
$$17+19+23=59$$
$$19+23+29=71$$
$$23+29+31=83$$
$$29+31+37=97$$
$$...$$
The number of such triplets, till a certain integer n, seems to be well approximated by the following function:
$$\frac{e\cdot\pi(n)}{\log(n)}$$



> 
> If the previous was true, the density of such primes in the whole set
>  of prime numbers would be comparable to that of primes inside the set
>  of naturals.
> 
> 
> **I ask if this estimate is correct and some references about this matter.**
> 
> 
> 


Thanks.



About the number of primes which are the sum of 3 consecutive primes (OEIS A034962)
The question asks how many primes $p\_n \le x$ are there such that $p\_n + p\_{n+1}+p\_{n+2}$ is also prime. This is beyond our reach to answer, but one can use Hardy-Littlewood type heuristics to attack this. Since $p\_n + p\_{n+1}+ p\_{n+2}$ is roughly of size $x$, it has about $1/\log x$ chance of being prime, and so one should expect the answer to be on the scale $\pi(x)/\log x \approx x/(\log x)^2$. But the asymptotic will be a little different, since $p\_n + p\_{n+1} +p\_{n+2}$ is not quite random -- for example it will always be odd (omitting $2+3+5$). 


Let's flesh out the usual heuristic. To be prime, a number must be coprime to each prime number $\ell$. A random integer has a chance $(1-1/\ell)$ of being coprime to $\ell$. What is the chance that $p\_n+ p\_{n+1} + p\_{n+2}$ is coprime to $\ell$? Each of the primes $p\_n$, $p\_{n+1}$, $p\_{n+2}$ itself lies in some reduced residue class $\mod \ell$. Assuming these possibilities are equally distributed, one should get the chance 
$$ 
\frac{1}{(\ell-1)^3} \sum\_{\substack{a, b, c=1 \\ (a+b+c,\ell)=1}}^{\ell-1} 1. 
$$ 
If $a$ and $b$ are given with $a\not\equiv -b \mod \ell$, then $c$ has $\ell-2$ possible residue classes, and if $a\equiv -b \mod \ell$ then $c$ has $\ell-1$ possible residue classes. So the answer is 
$$ 
\frac{1}{(\ell-1)^3} \Big( (\ell-1)(\ell-2)(\ell-2) + (\ell-1)(\ell-1)\Big)= 
\frac{\ell^2-3\ell+3}{(\ell-1)^2}. 
$$ 
So we adjust the random probability at $\ell$ by the factor 
$$ 
\Big(1-\frac {1}{\ell}\Big)^{-1} \frac{(\ell^2-3\ell+3)}{(\ell-1)^2}= 1 +\frac{1}{(\ell-1)^3}.
$$
For example, when $\ell=2$, this adjustment factor is $2$ reflecting the fact that $p\_n+p\_{n+1}+p\_{n+2}$ is guaranteed to be odd. 


Then the analog of the Hardy--Littlewood conjecture will be 
$$ 
\sim \prod\_{\ell} \Big(1+\frac{1}{(\ell-1)^3} \Big) \frac{x}{(\log x)^2}. 
$$ 
The constant in the product is approximately $2.3$; definitely not $e$. 


There are interesting biases concerning why this conjecture would be an underestimate. Work of [Lemke Oliver and Soundararajan](https://arxiv.org/abs/1603.03720) makes precise conjectures on the distribution of consecutive primes in arithmetic progressions. For example, if we look at $\ell=3$, the sum $p\_n + p\_{n+1} +p\_{n+2}$ can be a multiple of $3$ only if all three primes are congruent to each other $\mod 3$. This configuration is not preferred, and there are significant biases against it for small numbers! Similarly for other $\ell$ also, there are biases that indicate that the actual probability would be a tiny bit larger (with the effect wearing off slowly for large $x$). In other words, one might be able to identify lower order terms on the scale of $x(\log \log x)/(\log x)^3$, which might explain why the numerics suggest a larger value for the constant. 


