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 2019-12-20, 03:19 #1 miket   May 2013 916 Posts 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.