Does the constant 4.018 exists?

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.

• When $i \gt 1, a_i$ is primes of form $n^2 + (n+1)^2 + (n+2)^2, b^3 - b = (b-1)b(b+1)$. Dec 20, 2019 at 4:04

What you mean is, if $$b_n$$ is the $$n$$'th nonnegative integer $$j$$ such that $$3j^2+2$$ is prime, $$\lim_{n \to \infty} \dfrac{\sum_{j=1}^n (3 b_j^2+2)}{\sum_{j=1}^n (b_j^3-b_j)} b_n \approx 4.018$$ We don't even know for certain that there are infinitely many such integers, but heuristically it is likely that $$b_j \sim c j \log j$$ for some constant $$c$$ as $$j \to \infty$$. If so, $$\sum_{j=1}^n (3 b_j^2 + 2) \sim c^2 n^3 \log^2 n$$, $$\sum_{j=1}^n (b_j^3 - b_j) \sim c^3 n^4 \log^3(n)/4$$, and so your limit should be exactly $$4$$.