Let $n \in \mathbb{N}, n \geq 2$. By minimal Goldbach basis $G_{2n}$(if it is nonempty) of $2n$ , I mean the minimal set of primes such that every even number less than or equal to $2n$ can be written as a sum of two primes of that set. Minimal refers to the number of elements of the set. For instance,

$G_{4}= \left \{ 2 \right \}, G_{6}= \left \{ 2,3 \right \} , G_{8}= \left \{ 2,3,5 \right \}, G_{10}= \left \{ 2,3,5 \right \}$

However, $ \left \{ 2,3,5,7 \right \}$ also works for $ 2n = 10$ but it is not minimal.

The minimal set is unique since among many possibilities(if any) we are choosing one that has (1) the smallest number of elements(*this is the primary criterion*) and (2)say, if we have $\left \{ 2,3,5,11,13\right\}$ and $\left \{ 2,3,5,7,13\right\}$, we would choose the second. That means when ordering the elements in increasing order, the one chosen will have elements that are smaller under pairwise comparison with the other choices.

**Question 1:(not yet answered)** Obviously, $\left | G_{2p} \right |\leq \pi(p) $, for primes p. Proving that $\left | G_{2n} \right |= 0$ for at least one $n$ is harder than the old Goldbach. Still harder would be to show that $\left | G_{2n}\left | \leq \right |G_{2n+2} \right |$ Also, some experiment shows that
$\left | G_{2n} \right |= \pi(n) + 1$ when $n \geq6$ till the numbers I checked (excluding $n = 7$ and $13$). How good is this formula at least for some $n$?

**Question 2:(has been answered below)** Does $G_{2n}$ always contain the primes in linear order, without jumping any primes? (I asked this because none of the $G_{2n}$'s I computed have missing primes.)

Thanks.

**N.B.:** I have been making many revisions. So, some comments may not make sense. In that case, you may look up the edit history.

`$\left \{ 2,3,5,11,13\right\}$`

and`$\left \{ 2,3,5,7,13\right\}$`

, we would choose the second. That means when ordering the elements in increasing order, the one chosen will have elements that are smaller under pairwise comparison with the other choices. I will add this above. $\endgroup$ – Unknown Feb 8 '11 at 0:36