2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '19 at 4:04
6
$\begingroup$

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 $4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.