Primes of Atlantis: a definition and a problem

We denote the floor function as $\lfloor x\rfloor$, and for integers $k\geq 1$ we consider the following sum of the areas of three consecutive circles $$\pi k^2+\pi(k+1)^2+\pi(k+2)^2=\pi \left( 3k^2+6k+5\right).$$

Definition. When for an integer $n\geq 1$ the integer $$\mathcal{A}(n)=\lfloor \pi\left( 3n^2+6n+5\right) \rfloor$$ is a prime number, I say that it's a prime of Atlantis.

Our sequence of prime of Atlantis starts as $$43, 157, 241, 769, 4567, 11551, 14341, 16631, 19949\ldots$$ corresponding to the indexes $n's:1, 3, 4, 8, 21, 34, 38, 41, 45, \ldots$ as you can see with these codes using Wolfram Alpha online calculator:

Table IsPrime(floor(pi (n^2+(n+1)^2+(n+2)^2))), for n=1 to 100

Table floor(pi (n^2+(n+1)^2+(n+2)^2)), for n=1 to 100

Question. I would like to know if we can deduce if there are infinitely many primes of Atlantis. Many thanks.

If you can't solve the problem, but you can provide us useful reasonings or calculations about the Question, please share your knowledges.

• Many thanks for your attention. Now I don't understand well your question. I don't know if your question is related with my calculations. On the other hand notice that I am a bad programmer @Peter When I wrote my previous codes that I evoke is that one can do a comparison of both tables. – user243301 Sep 30 '17 at 12:24
• For $n=10^{100}+142$, we get a large "prime of Atlantis". There should be infinite many, but proves that a sequence contains infinite many primes , in particular if irrational numbers and floor functions are involved, are usually out of reach. – Peter Sep 30 '17 at 12:24
• What I mean : You used Wolfram Alpha and apparently got only the array with the integers. What was your exact input ? – Peter Sep 30 '17 at 12:25
• About your opinion I am agree since this kind of problems are (miscellaneous) difficult to solve. I did this question to try some useful reasoning (see my last paragraph). On the other hand your prime $10^{100}+142$ sems fantastic and ...is the largest number known, so far! Many thanks for share it @Peter – user243301 Sep 30 '17 at 12:28
• The number-theoretic tools available are almost certainly not enough to solve this question. We don't even know the answer to a seemingly (slightly) simpler problem - whether there are infinitely primes of the form $n^2+1$. Throwing transcendental number into the problem is probably not going to make it simpler... – Wojowu Oct 8 '17 at 17:28

This is not a complete answer , but I show some statistics about the "primes of Atlantis"

Program PARI/GP is used

First of all the $n$ upto $1000$, for which we get a "prime of Atlantis" :

\p 10000 { High precision calculation }
pa(n)=isprime(truncate(Pi*(3*n^2+6*n+5)),2)==1 {pa(n) is true if and only if we get a prime of Atlantis by choosing $n$}

? select(m->pa(m)==1,[0..1000])
%4 = [1, 3, 4, 8, 21, 34, 38, 41, 45, 69, 75, 91, 136, 166, 179, 190, 202, 222,
227, 228, 229, 230, 239, 267, 284, 308, 313, 317, 323, 351, 359, 381, 392, 409,
417, 429, 433, 434, 442, 449, 456, 460, 463, 468, 486, 490, 518, 527, 549, 585,
588, 607, 632, 651, 668, 670, 684, 694, 700, 703, 727, 764, 775, 782, 805, 811,
814, 820, 821, 844, 850, 894, 896, 920, 925, 926, 932, 969, 979, 985]
?


For $n\in [1,10]$ , we get $4$ primes

For $n\in [10,100]$ , we get $8$ primes

For $n\in [100,1000]$ , we get $68$ primes

For $n\in [10^3,10^4]$ , we get $498$ primes

For $n\in [10^4,10^5]$ , we get $3778$ primes

For $n\in [10^5,10^6]$ , we get $31767$ primes

Some large $n$, for which we get a "prime of Atlantis"

$$10^{100}+142$$ $$10^{200}+114$$ $$10^{500}+6391$$ $$10^{1000}+1395$$

with $201$ , $401$ , $1001$ and $2001$ digits respectively. There are probably infinite many "primes of Atlantis", but it is hard to imagine that this can be proven.

• Now without jokes in my words: your work is incredible, thank you very much for your statistics and conjecture. A vote up, and if I could, I should vote your answer 101 times. – user243301 Sep 30 '17 at 15:32
• @user243301 Thank you, at least you can accept it ... – Peter Sep 30 '17 at 16:00