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%I #31 Sep 08 2022 08:44:53
%S 2,3,5,11,17,23,29,41,47,53,59,61,67,71,73,83,89,101,103,107,113,131,
%T 137,149,151,167,173,179,191,193,197,227,233,239,251,257,263,269,271,
%U 281,293,307,311,317,347,353,359,367,383,389,401,419,431,439,443,449,461,467,479,491,499
%N Primes p such that x^3 = 3 has a solution mod p.
%C Complement of A040038 relative to A000040. - _Vincenzo Librandi_, Sep 13 2012
%C Being a cube mod p is a necessary condition for 3 to be a 9th power mod p. See Williams link pp. 1, 8 (warning: term 271 is missed). - _Michel Marcus_, Nov 12 2017
%H Vincenzo Librandi, <a href="/A040036/b040036.txt">Table of n, a(n) for n = 1..1000</a>
%H Kenneth S. Williams, <a href="http://www.mscand.dk/article/view/11555/9571">3 as a Ninth Power (mod p)</a>, Math. Scand. Vol 35 (1974), 309-317.
%p select(p -> isprime(p) and numtheory:-mroot(3,3,p) <> FAIL, [2,seq(i,i=3..1000,2)]); # _Robert Israel_, Nov 12 2017
%t ok [p_]:=Reduce[Mod[x^3 - 3, p] == 0, x, Integers] =!= False; Select[Prime[Range[200]], ok] (* _Vincenzo Librandi_, Sep 11 2012 *)
%o (Magma) [p: p in PrimesUpTo(450) | exists(t){x : x in ResidueClassRing(p) | x^3 eq 3}]; // _Vincenzo Librandi_, Sep 11 2012
%o (PARI) isok(p) = isprime(p) && ispower(Mod(3,p), 3); \\ _Michel Marcus_, Nov 12 2017
%Y Cf. A000040, A040038. Contains A003627.
%K nonn,easy
%O 1,1
%A _N. J. A. Sloane_
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