1: Prime gap bounds:
Consider the following non-asymptotic bounds for $\pi(x)$, proven by Dusart in 2018 (holding for $x>5393$):$$\frac{x}{\log(x)-1}<\pi(x)<\frac{x}{\log(x)-1.112}$$
To get the maximum value for $p(n)$, we take the lower bound with $x=p(n)$ :
$$\pi(p(n))=\frac{p(n)}{\log p(n)-1}$$ $$n\left(\log p(n)-1\right)-p(n)=0$$ $$p(n)=-nW_{-1}\left(\frac{-e^{1}}{n}\right)$$
$W_{k}(n)$ is the analytic continuation of the product log function. Now if we do the same with the upper bound, we get non-asymptotic bounds for the $N^{th}$ prime $p(n)$:
$$-nW_{-1}\left(\frac{-e^{1.112}}{n}\right)<p(n)<-nW_{-1}\left(\frac{-e}{n}\right)$$
Using our bounds for $p(n)$, we can get a upper bound for the $n^{th}$ prime gap $g(n)$. Since $g(n)=p(n+1)-p(n)$, we will take the upper bound for $p(n+1)$ and the lower bound for $p(n)$:
$$2 \leq g(n)< \left(-(n+1)W_{-1}\left(\frac{-e}{(n+1)}\right)\right)-\left(-nW_{-1}\left(\frac{-e^{1.112}}{n}\right)\right)$$
2: Goldbach critical prime gap:
Let's define $a(n)$, the highest even integer such that all positive even integer $\leq a(n)$ except $2$ and $4$ can be written as the sum of $2$ odd prime $\leq p(n)$. We can now reformulate Goldbach's conjecture as:$$a(n)\geq p(n), 1<n<\infty$$
Now since we know all even integer from $p(n)$ to $a(n)$ already have at least $1$ Goldbach partition, we can define $c(n)=a(n)-p(n)$ as the Goldbach critical prime gap, i.e the minimum value of the prime gap $g(n)$ such that an even integer between $a(n)$ and $p(n+1)$ have no Goldbach partition.
Note: $a(n)=$A090259$-2$.
3: Observations:
Two observations that seems to makes it impossible for Goldbach's conjecture to be false are:$$\lim_{n\rightarrow\infty}\frac{p(n)}{a(n)}\approx\frac{1}{2}$$
$2^{1}$ | $0.50000$ $2^{10}$ | $0.51310$
$2^{2}$ | $0.53846$ $2^{11}$ | $0.50561$
$2^{3}$ | $0.51351$ $2^{12}$ | $0.50168$
$2^{4}$ | $0.59550$ $2^{13}$ | $0.50159$
$2^{5}$ | $0.54811$ $2^{14}$ | $0.50102$
$2^{6}$ | $0.54657$ $2^{15}$ | $0.50052$
$2^{7}$ | $0.53062$ $2^{16}$ | $0.50031$
$2^{8}$ | $0.52208$ $2^{17}$ | $0.50015$
$2^{9}$ | $0.51711$ $2^{18}$ | $0.50009$
$$\lim_{n\rightarrow\infty}\frac{\left(-(n+1)W_{-1}\left(\frac{-e}{(n+1)}\right)\right)-\left(-nW_{-1}\left(\frac{-e^{1.112}}{n}\right)\right)}{c(n)}\approx0.112$$
$2^{1}$ | $0.00000$ $2^{10}$ | $0.13222$
$2^{2}$ | $1.93764$ $2^{11}$ | $0.12654$
$2^{3}$ | $0.43229$ $2^{12}$ | $0.12335$
$2^{4}$ | $0.37365$ $2^{13}$ | $0.12240$
$2^{5}$ | $0.22899$ $2^{14}$ | $0.12142$
$2^{6}$ | $0.18954$ $2^{15}$ | $0.12059$
$2^{7}$ | $0.15974$ $2^{16}$ | $0.12000$
$2^{8}$ | $0.14525$ $2^{17}$ | $0.11950$
$2^{9}$ | $0.13731$ $2^{18}$ | $0.11909$
