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%I #33 Sep 08 2022 08:44:53
%S 7,13,19,37,61,67,73,79,97,103,139,151,163,181,193,199,211,241,271,
%T 313,331,337,349,367,373,379,409,421,463,487,523,541,547,571,577,607,
%U 613,619,631,661,673,709,751,757,769,787,823,829,853,859,877,883,907,937
%N Primes p such that x^3 = 2 has no solution mod p.
%C Primes represented by the quadratic form 4x^2 + 2xy + 7y^2, whose discriminant is -108. - _T. D. Noe_, May 17 2005
%C Complement of A040028 relative to A000040. - _Vincenzo Librandi_, Sep 17 2012
%H Klaus Brockhaus, <a href="/A040034/b040034.txt">Table of n, a(n) for n=1..1000</a>
%H Steven R. Finch, <a href="http://arXiv.org/abs/math.NT/0701251">Powers of Euler's q-Series</a>, (arXiv:math.NT/0701251), 2007.
%e A cube modulo 7 can only be 0, 1 or 6, but not 2, hence the prime 7 is in the sequence.
%e Because x^3 = 2 mod 11 when x = 7 mod 11, the prime 11 is not in the sequence.
%t insolublePrimeQ[p_]:= Reduce[Mod[x^3 - 2, p] == 0, x, Integers] == False; Select[Prime[Range[200]], insolublePrimeQ] (* _Vincenzo Librandi_ Sep 17 2012 *)
%o (Magma) [ p: p in PrimesUpTo(937) | forall(t){x : x in ResidueClassRing(p) | x^3 ne 2} ]; // _Klaus Brockhaus_, Dec 05 2008
%o (PARI) forprime(p=2,10^3,if(#polrootsmod(x^3-2,p)==0,print1(p,", "))) \\ _Joerg Arndt_, Jul 16 2015
%K nonn,easy
%O 1,1
%A _N. J. A. Sloane_
%E More terms from _Klaus Brockhaus_, Dec 05 2008
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