# What is the probability that $\exists N \in \mathbb{N}$ such that $\forall n > N$, $2n \in C + C$?

Suppose $$C$$ is a random subset of $$\mathbb{N}\setminus\{1, 2\}$$, such that $$\forall n \in \mathbb{N}\setminus\{1, 2\}$$, $$P(n \in C) = \frac{1}{\ln(n)}$$ and the events of different numbers belonging to $$C$$ are independent (Such random subset is called Cramer’s Prime Model, as it was first introduced by Harald Cramer as probabilistic model for distribution of prime numbers). What is the probability that $$\exists N \in \mathbb{N}$$ such that $$\forall n > N$$, $$2n \in C + C$$, where $$+$$ stands for pointwise summation?

This question was inspired by the famous Goldbach conjecture from number theory:

$$\exists N \in \mathbb{N}\,|\,\forall n > N$$, $$2n \in P + P$$, where $$P$$ is the set of all prime numbers.

That conjecture was first stated in 1742, remains open until now and is considered to be quite hopeless. However, maybe, it is known, what is the probability of similar statement being true in Cramers Model…

• Perhaps you may want to remove even numbers from $C$. Any case, even if the probability is $1$, this would not prove the Goldbach's conjecture. – ajotatxe Apr 16 at 12:04
• @ajotatxe, yes, it will not. I never stated, that it would... – Yanior Weg Apr 16 at 12:10

This probability is easy to find. For any $$k$$ from $$3$$ to $$n-1$$ the probability that $$k$$ or $$2n-k$$ are not in $$C$$ is $$1-\frac1{\ln k\ln(2n-k)}$$ And the probability, that $$n$$ is not in $$C$$ is $$1-\frac1{\ln n}$$ So $$p(2n\in C+C)=1-\left(1-\frac1{\ln n}\right)\prod_{k=3}^{n-1}\left(1-\frac1{\ln k\ln(2n-k)}\right)$$
$$n^{-2} \ge e^{-\frac{n}{2\log^2 n}} \ge \left(1-\frac1{\ln n}\right)\prod_{k=3}^{n-1}\left(1-\frac1{\ln k\ln(2n-k)}\right)$$
As $$\sum_{n \in \mathbb{N}} \frac{1}{n^2}$$ is finite, it follows from 1st Borel-Cantelli Lemma that the probability that there exists an infinite number of integers $$n$$ such that $$2n \not \in C+C$$ is 0. So the probability that there exists an $$N$$ as in the problem statement is 1.