# A heuristic argument for the Goldbach conjecture?

This question here is purely speculative so be warned if you read on: This question is related to a sequence $$b_n$$ which is defined here:

A series related to prime numbers

For the numbers $$a_{2n,2}$$ of ordered sums of writing $$2n$$ as a sum of two primes, we have:

$$a_{2n,2} = \frac{1}{n-2} \sum_{v=0}^{2n-1} a_{v,2} \cdot b_{2n-1-v}$$

which using the conjecture $$b_n / b_{n+1} \approx \gamma = -0.62923367 \cdots$$ becomes:

$$\approx \frac{1}{n-2} \sum_{v=0}^{2n-1} a_{v,2} \gamma^v b_{2n-1}$$

We can divide this last sum into two parts:

$$=\frac{b_{2n-1}}{n-2}( \sum_{v=0,v \equiv 0 (2)}^{2n-1} a_{v,2} \gamma^v + \sum_{v=0,v\equiv 1 (2)}^{2n-1} a_{v,2} \gamma^v )$$

For the second sum observe, that $$a_{v,2} = 2$$ if and only if $$v=p+2$$ for some prime $$p$$. Hence we get for the second sum, shoul be :

$$\sum_{v=0,v\equiv 1 (2)}^{2n-1} a_{v,2} \gamma^v = 2 \sum_{v=p+2, p \text{prime}}^{2n-1} \gamma^{p+2}$$ Since for large $$n$$ we have $$2 f(t)\cdot t^2 \approx 2 \sum_{p we get for the second sum ,since ( as remains to be shown) $$f(\gamma)=0$$: $$\approx 2 f(\gamma)\cdot \gamma^2 = 0$$.

The first sum is by induction on $$n$$ ( $$a_{v,2} \ge 1$$ for $$v \equiv 0 (2), v < 2n-1$$ ) not zero. Hence the whole sum should not be zero.

One thing to observe for this argumentation is $$b_{2n-1}$$ seems to always be negative, so since this is only a heuristic, this should not be a problem.

My questions is, if you can think of some way to make the steps above more rigorous, if this is not asked to much?