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-   -   Does the constant 4.018 exists? (https://www.mersenneforum.org/showthread.php?t=25038)

miket 2019-12-20 03:19

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.

miket 2019-12-20 06:53

Robert Israel answer this question at math.stackexchange [URL="https://math.stackexchange.com/questions/3482438/does-the-constant-4-018-exists"]Does the constant 4.018 exists?[/URL]:

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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