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%I #41 Aug 20 2022 18:44:26
%S 1,5,1,1,5,1,13,5,17,19,1,11,1,5,1,29,31,1,17,1,19,13,41,43,1,23,1,1,
%T 17,53,1,19,29,1,31,1,13,67,23,71,1,37,1,1,79,1,83,1,43,1,1,13,31,1,1,
%U 1,1,1,103,1,107,109,1,1,1,29,1,1,1,61,1,1
%N a(n) = b(n+1)/b(n) - 1 where b(1)=3 and b(k) = b(k-1) + lcm(floor(sqrt(3)*k), b(k-1)).
%C Conjectures:
%C 1. This sequence consists only of 1's and primes.
%C 2. Every odd prime of the form floor(sqrt(3)*m) greater than 3 is a term of this sequence.
%C 3. At the first appearance of each prime of the form floor(sqrt(3)*m), it is larger than any prime that has already appeared.
%C The 2nd and 3rd conjectures are proved at the Mathematics Stack Exchange link. - _Sungjin Kim_, Jul 17 2019
%H Michel Marcus, <a href="/A323388/b323388.txt">Table of n, a(n) for n = 1..10000</a>
%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/3290665/generating-prime-numbers-of-the-form-lfloor-sqrt3-cdot-n-rfloor">Generating primes of the form floor(sqrt(3)*n)</a>
%o (PARI) Generator(n)={b1=3; list=[]; for(k=2, n, b2=b1+lcm(floor(sqrt(3)*k), b1); a=b2/b1-1; list=concat(list,a); b1=b2); print(list)}
%o (PARI) lista(nn)={my(b1=3, b2, va=vector(nn)); for(k=2, nn+1, b2=b1+lcm(sqrtint(3*k^2), b1); va[k-1]=b2/b1-1; b1=b2); va}; \\ _Michel Marcus_, Aug 20 2022
%Y Cf. A184796 (primes of the form floor(sqrt(3)*m)).
%Y Cf. A135506, A323359, A323386.
%K nonn
%O 1,2
%A _Pedja Terzic_, Jan 13 2019
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