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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 $2 f(t)\cdot t^2 \approx 2 \sum_{p<n, p \text{ prime }} t^{p+2}$ 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?

Thanks for your help!

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