Every sufficiently large positive integer is the average of $n$ distinct primes for certain $n \geq 2$?

Question

I want to generalize a stronger Goldbach's conjecture a little bit because that might help solve it.

I was thinking:

For all $n \geq 2$, every sufficiently large positive integer $x \geq b_n$ is the average of $n$ distinct primes?

Clearly this implies Goldbach's conjecture.

So, was wondering if it has already been proven for any $n\gt 2$?

I think Terry Tao proved something different to the above, since it explicitly says at most 5 primes, which is algebraically not the same thing.

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Remark. In particular the bound $b_n \geq \dfrac{p_1 + \dots + p_n}{n}$, where $p_i$ is the $i$th prime number: $2, 3, 5, \dots $

Proof of remark. If $\dfrac{q_1 + \dots +q_n}{n} = x \lt \dfrac{\sum_{i=1}^n p_i}{n}$, then cancel out the $n$ and the result follows. The $q_i$ are some prime numbers averaging to $x$.

For example for $n = 2$, the lower bound of $x$, $b_n$ seems to be $b_n = 4$, where $\dfrac{3 + 5}{2} = 4$ shows a solution for $x = 4$.

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Here are some small cases:

$$ n = 2: \\ \dfrac{5 + 3}{2} = 4 \\ \dfrac{7 + 3}{2} = 5 \\ \dfrac{5 + 7}{2} = 6 \\ \dfrac{11 + 3}{2} = 7 \\ \dfrac{11 + 5}{2} = 8 \\ \dfrac{11 + 7}{2} = 9 \\ \ \\ n = 3: \\ \dfrac{3 + 5 + 7}{3} = 5 \\ \dfrac{2 + 5 + 11}{3} = 6 \\ \ \\ n = 4: \\ \dfrac{3 + 5 + 7 + 13}{4} = 7\\ \vdots $$

We will need some SymPy Python 3.x code. I might code it up (testing up to $x = N, n = M$), but not for a few days. So there is some opportunity to beat me to it.

Answer 1

For $n$ large enough Vinogradov's theorem gives $\sim C \frac{n^2}{\ln^3 n}$ solutions to $2n+1 = p_1+p_2+p_3$ which gives $\sim B \frac{n^3}{\ln^4 n}$ solutions to $2n = p_1+p_2+p_3+p_4$.

And the same method as Vinogradov gives $\sim A \frac{n^2}{\ln^3 n}$ solutions to $2n = 2q_1+q_2+q_3$.

Thus we have $\sim B \frac{n^3}{\ln^4 n}$ solutions for $2n = p_1+p_2+p_3+p_4$ as the sum of distinct primes.

Vinogradov's theorem follows from a strong form of the PNT in arithmetic progressions. I am quite sure for large enough $k$ there is a three line proof to your $2n = \sum_{j=1}^k p_j$ distinct primes problem just from $\pi(x) \sim \frac{x}{\ln x}$.

Answer 2

Your conjecture it's not new, Hardy and Littlewood in 1923 conjectured that for a fixed $c \in \mathbb{N} \geq 2$ and large $n \in \mathbb{N}$:

The number of representations of $n$ as the sum of $c$ primes $n=p_1+p_2+\cdots+p_c$ with $p_1 \leq p_2 \leq \cdots \leq p_c$ is given asymptotically by :

$$\left( \prod_{\text{p prime}} \dfrac{p \, \gamma_{c,p}(n)}{(p-1)^c} \right) \int_{2 \leq x_1 \leq \cdots \leq x_c; n=x_1+x_2+\cdots+x_c} \dfrac{dx_1 \cdots dx_c}{\log(x_1) \cdots \log(x_c)}$$

With $\gamma_{c,p}(n)$ is the number of solutions to the equation $n=q_1+q_2+\cdots+q_c \pmod p$ with $q_1,q_2,\cdots q_c \neq 0 \pmod p$

This conjecture was proven to be true for $c \geq 3$ by Vinogradov, but the case $c=2$ still open (Goldbach's conjecture).

In your conjecture we take $m=n \, c$.

More details : here