

A003630


Inert rational primes in Q(sqrt(3)).
(Formerly M3766)


4



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

1,1


COMMENTS

Primes p such that p divides 3^(p1)/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
Conjecture: Primes congruent to (5, 7) mod 12.  Vincenzo Librandi, Aug 06 2012
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


REFERENCES

H. Hasse, Number Theory, SpringerVerlag, NY, 1980, p. 498.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe, Table of n, a(n) for n = 1..1000


EXAMPLE

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.


MATHEMATICA

Select[Prime[Range[2, 200]], JacobiSymbol[3, #] == 1 &] (* Alonso del Arte, Mar 26 2017 *)


PROG

(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 */


CROSSREFS

Cf. A010527.
Sequence in context: A099382 A163633 A092242 * A122565 A247607 A079016
Adjacent sequences: A003627 A003628 A003629 * A003631 A003632 A003633


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane, Mira Bernstein


STATUS

approved



