Does the constant 4.018 exists?

miket · 2019-12-20, 03:19

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.

miket · 2019-12-20, 06:53

Robert Israel answer this question at math.stackexchange Does the constant 4.018 exists?:

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 \(4\).

2019-12-20, 03:19	#1
miket May 2013 3² 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.

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2019-12-20, 06:53	#2
miket May 2013 3² Posts	Robert Israel answer this question at math.stackexchange Does the constant 4.018 exists?: 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 \(4\).