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A003630
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Inert rational primes in Q[sqrt(3)].
(Formerly M3766)
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8
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5, 7, 17, 19, 29, 31, 41, 43, 53, 67, 79, 89, 101, 103, 113, 127, 137, 139, 149, 151, 163, 173, 197, 199, 211, 223, 233, 257, 269, 271, 281, 283, 293, 307, 317, 331, 353, 367, 379, 389, 401, 439, 449, 461, 463, 487, 499, 509, 521, 523, 547, 557, 569, 571, 593
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OFFSET
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1,1
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COMMENTS
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Primes p such that p divides 3^(p-1)/2 + 1. - Cino Hilliard, Sep 04 2004
Primes p such that 1 + 4*x + x^2 is irreducible over GF(p). - Joerg Arndt, Aug 10 2011
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
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
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REFERENCES
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H. Hasse, Number Theory, Springer-Verlag, NY, 1980, p. 498.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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EXAMPLE
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Since (-1)*(1 - sqrt(3))*(1 + sqrt(3)) = 2, 2 is not in the sequence.
3 is not in the sequence for obvious reasons.
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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MATHEMATICA
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Select[Prime[Range[2, 200]], JacobiSymbol[3, #] == -1 &] (* Alonso del Arte, Mar 26 2017 *)
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PROG
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(PARI) {a(n) = local( cnt, m ); if( n<1, return( 0 )); while( cnt < n, if( isprime( m++) && kronecker( 12, m )== -1, cnt++ )); m} /* Michael Somos, Aug 14 2012 */
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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