TITLE: Finite sums of prime numbers $\geq x$
QUESTION [19 upvotes]: Let $S_x$ be the set of finite sums of prime numbers $\geq x$. In other words, let $S_x$ be the submonoid of $(\mathbf{Z}_{\geq 0},+)$ generated by the set $\mathcal{P}_{\geq x}$ of prime numbers $\geq x$.
It is easy to see that $S_x$ contains every sufficiently large integer. This follows from the classical fact that given two coprime integers $a$ and $b$, every sufficiently large integer, in fact every integer $\geq (a-1)(b-1)$, is of the form $ma+nb$ for some $m, n \in \mathbf{Z}_{\geq 0}$. See for example this page.
Let $N_x$ be the largest integer which is not in $S_x$.
Examples :
If $x=2$ then $S_2 = \{0\} \cup \mathbf{N}_{\geq 2}$ so that $N_2=1$.
If $x=3$ then $S_3 = \{0,3\} \cup \mathbf{N}_{\geq 5}$ so that $N_3=4$.
By definition, we have $N_x \geq x-1$ (in fact parity considerations imply that $N_x \geq 2x-2$ for $x \geq 3$).
On the other hand, given that there are at least two primes in the interval $[x,2x]$, the above classical fact implies that $N_x \ll x^2$.

What is the asymptotic behaviour of $N_x$ as $x \to \infty$ ?

REPLY [4 votes]: According to "The three primes theorem with almost equal summands" by Baker and Harman, every large odd $N$ is a sum of three primes each of size $\sim N/3$. (In fact, within $N^{4/7}$ of $N/3$-- this is much closer than we need, so we could use weaker results of earlier authors, if preferred.) 
For odd $N$, this gives $N$ as a sum of three primes, each at least $\sim N/3$. If $N$ is even, take a prime $p$ of size $\sim N/4$ (which exists by the prime number theorem), and write $N-p$ as a sum of three primes of size about $N/4$. So we get $N$ as a sum of four primes at least $\sim N/4$.