Let $A$ be the set $\{a_1,a_2,\ldots,a_n\},$ for each $i, a_i $is prime number of the form $3j^2+2, j \geq 0 $
let $B$ be the set $\{b_1,b_2,\ldots,b_n\}$, for each $i, 3b_i^2+2$ is prime number,$ b_i \geq 0 $
Let $$ f(n)=\frac{\quad\sum A}{\quad\sum_{b\in B} b^3 - b}b_n, b \in B$$
For example, when $n=3$, $$f(3)= \frac{2+5+29}{0^3 - 0 + 1^3 - 1 + 3^3 - 3} \times 3 = 4.5 $$
When $n=40400$, $$f(40400)=\dfrac{38237010330695965}{9515800255043913608016} \times 999967 \approx 4.018 $$
When $n=2988619$, $$f(2988619)=\dfrac{28727312822972002780844}{714881028260333643707250890088} \times 99999987 \approx 4.018 $$
Is it possible that $$\lim_{n\to+\infty}f(n) \approx 4.018?$$
I only check $b_n$ to $10^8$, furthermore check are welcome.