I'm interested in all natural numbers, that are the sum of pariwise different prime numbers. For example 10 is such a number, since
$10 = 7 + 3$
My suspicion is that almost every natural number is the sum of pairwise different prime numbers. Actually by my pen & paper calculations, I did not find a natural number larger than 6 that is not such a sum, but I did not calculate much, so maybe there are even small numbers that are not a sum of pairwise different primes.
So my questions are these:
Is there any natural number larger than 6, which is not the sum of pairwise different primes? If not, any idea how to prove this?
And what's about the natural numbers, that are the sum of at least n natural prime numbers, with $n \in \mathbb{N}$? Is there any knowledge about these numbers?
Regards,
S. M. Roch
Edit: I'm sorry, I forgot to say for my first question, that I also accept the sum of not only two prime numbers. So for example
$20 = 11 + 7 + 2$
is also such a sum, and for every prime number the number is its sum itself.
Best Answer
A sketch of a proof:
For sufficiently large $N$, the Prime Number Theorem guarantees a prime $p$ between $N/2$ and $N-6$, $N/2\lt p\lt N-6$. This implies $6\lt N-p\lt N/2\lt N$, so strong induction tells us $N-p$ is the sum of distinct primes, $N-p=p_1+p_2+\cdots+p_n$ with $p_1\lt p_2\lt\cdots\lt p_n$. But $N-p\lt N/2\lt p$ tells us these primes are all less than $p$. Thus $N=p_1+p_2+\cdots+p_n+p$ with $p_1\lt p_2\lt\cdots\lt p_n\lt p$.
Note, Bertrand's postulate, which guarantees a prime between $N/2$ and $N$, isn't quite good enough. E.g., the only primes between $17/2$ and $17$ are $11$ and $13$, and neither $17-11=6$ nor $17-13=4$ can be written as a sum of distinct primes.