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A172490 Primes p of the form 4m+3 for which there are exactly as many primitive roots modulo p in the interval [0,p/2] as in the interval [p/2,p].

%I A172490 #18 Nov 23 2019 23:04:58
%S A172490 7,31,43,67,307,367,487,643,1327,1663,2371,3643,3847,4327,4951,6091,
%T A172490 6571,8263,9151,9187,11239,11383,11863,15307,24007,24151,27847,30091,
%U A172490 30643,33619,36871,42187,44171,46279,46591,48787,70843,71887,72103,72379,73363,79867,82003,92503,95467,106243,110431,120943,126031,130363,139759,143827,162751,167107,173191,174859,183247
%N A172490 Primes p of the form 4m+3 for which there are exactly as many primitive roots modulo p in the interval [0,p/2] as in the interval [p/2,p].
%C A172490 Primes 4*k+3 where half of the primitive roots are <= (p-1)/2.
%C A172490 The sequence is probably infinite.
%C A172490 Primes of the form 4m+1 always have as many primitive roots in [0,p/2] as in [p/2,p] (see A172480).
%p A172490 with(numtheory): p:=3: while p<1000 do if(p mod 4 = 3)then b1:=0: b2:=0: m:=primroot(p): while not m=FAIL do if(m<p/2)then b1:=b1+1: else b2:=b2+1: fi: m:=primroot(m,p): od: if(b1=b2)then printf("%d, ",p): fi: fi: p:=nextprime(p): od: # _Nathaniel Johnston_, Jun 26 2011
%t A172490 << NumberTheory`NumberTheoryFunctions` m = 2; s = {}; While[m < 10000, m++; p = Prime[m]; If[Mod[p, 4] == 1, , q = (p - 1)/2; g = PrimitiveRoot[p]; se = Select[Range[p - 1], GCD[ #, p - 1] == 1 &]; e = Length[se]; j = 0; t = 0; While[j < e, j++; h = PowerMod[g, se[[j]], p]; If[h <= q, t = t + 1,] ]; If[e == 2t, s = {s, p},] ] ]; s = Flatten[s]
%o A172490 (PARI) isA172490(p)=isprime(p)&&p%4==3&&sum(n=0,p\2,gcd(n,p)==1&&znorder(Mod(n,p))==p-1)==sum(n=p-p\2,p,gcd(n,p)==1&&znorder(Mod(n,p))==p-1) \\ _Charles R Greathouse IV_, Jun 27 2011
%Y A172490 Cf. A118818, A172480.
%K A172490 hard,nonn
%O A172490 1,1
%A A172490 _Emmanuel Vantieghem_, Feb 05 2010
%E A172490 More terms from _Robert Israel_, Nov 23 2019