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<a href="https://oeis.org/index/Pri#primes_decomp_of">Index to sequences related to decomposition of primes in quadratic fields</a>
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<a href="https://oeis.org/wikiindex/Index_to_OEIS:_Section_Pri
Since (-1)*(1 - sqrt(3))*(1 + sqrt(3)) = 2, 2 is not in the sequence.
x^2 = = 3 (mod 5 ) has no solution, which means that 5 is an inert prime in Z[sqrt(3)]. Therefore, 5 is in the sequence.
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The above conjecture is correct. In fact, this is the sequence of primes p such that Kronecker(12,p) = -1 (12 is the discriminant of Q[sqrt(3)]), that is, odd primes that have 3 as a quadratic nonresidue. - Jianing Song, Nov 21 2018
The conjecture from Vincenzo Librandi is correct. In fact, this sequence is the primes p such that Kronecker(12,p) = -1 (12 is the discriminant of Q[sqrt(3)]), that is, odd primes that have 3 as a quadratic nonresidue. - Jianing Song, Nov 21 2018
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Conjecture: Let r(n) = (a(n) - 1)/(a(n) + 1)) if a(n) mod 4 = 1, (a(n) + 1)/(a(n) - 1)) otherwise; then Product_{n>=1} r(n) = (2/3) * (4/3) * (8/9) * (10/9) * (14/15) * ... = sqrt(3)/2. (See A010527.) We see that the sum of the numerator and denominator of each fraction equals the corresponding term of the sequence: 2 + 3 = 5, 4 + 3 = 7, 8 + 9 = 17, ... - Dimitris Valianatos, Mar 26 2017