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](http://en.wikipedia.org/wiki/Coin_problem).


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 $N\_x \ll x^2$.



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> What is the asymptotic behaviour of $N\_x$ as $x \to \infty$ ?
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Finite sums of prime numbers $\geq x$
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 $\sim N/4$. 


