I explained in another question how I got to this formula for $g(n)$, the number of Goldbach partitions of even integer $2n$:
$$g_{\left(n\right)}=\sum_{p\leq2n-3}\omega\left(2n-p\right)-\sum_{p\leq2n-3}\pi_{p,b}\left(2n-3p\right)$$
where $p$ is odd prime, $\pi_{a,b}\left(n\right)$ denotes the number of primes of the form $ak+b$ less than or equal to $n$ and:
$$b=\left(2n-3p\right)\mod p$$
Now I used the following asymptotic relations:
$$\omega\left(n\right)\sim\log\log n$$ $$\pi_{p,b}\left(n\right)\sim\frac{\pi\left(n\right)}{\varphi\left(p\right)}\sim\frac{\pi\left(n\right)}{p-1}$$
which gives me this:
$$g_{\left(n\right)}\sim\sum_{2<p\leq2n-3}\log\log\left(2n-p\right)-\sum_{2<p\leq2n-3}\frac{\pi\left(2n-3p\right)}{p-1}$$
Now we know that if $p$ is a prime factor of $n$, then $\pi_{p,b}\left(2n-3p\right)=1$, so:
$$\sum_{_{2<p\leq2n-3}}\frac{\pi\left(2n-3p\right)}{p-1}$$
would be the maximum value for: $$\sum_{_{p\leq2n-3}}\pi_{p,b}\left(2n-3p\right)$$
and will occur when $n$ have no prime factor $\leq\frac{2n-3}{3}$. In fact, a very accurate estimation is:
$$\sum_{2<p\leq2n-3}\pi_{p,b}\left(2n-3p\right)\sim\sum_{2<p\leq2n-3}\frac{\pi\left(2n-3p\right)}{p-1}-\sum_{p>2,p\mid n}\frac{\pi\left(2n-3p\right)}{p-1}$$
Question 1:
When we look at the graph for: $$\sum_{2<p\leq2n-3}\pi_{p,b}\left(2n-3p\right)$$
in red, and
$$ \sum_{2<p\leq2n-3}\log\log\left(2n-p\right) $$
in green:
It looks like the following holds for large $n$: $$\sum_{2<p\leq2n-3}\pi_{p,b}\left(2n-3p\right)<\sum_{2<p\leq2n-3}\log\log\left(2n-p\right)$$
I computed up to $10^7$ and it holds for all $n>41011$
Does anyone have an idea on how to prove this?
Question 2:
Looking at the asymptotic result for $g(n)$, It also looks like a lower bound:
$$ g\left(n\right)>\sum_{2<p\leq2n-3}\log\log\left(2n-p\right)-\sum_{2<p\leq2n-3}\frac{\pi\left(2n-3p\right)}{p-1}$$
This one holds for all $n>2$. Of course this one should be harder to prove because it would prove Goldbach's conjecture. Any idea on how it could or could not be possible to prove those bounds ?
Also, I came up with this formula myself, but I haven't found anything similar on the web yet, so it would also be helpful if someone could think of some related work I could read.