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.

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

1 Answer 1


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 $4$.


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