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].

7, 31, 43, 67, 307, 367, 487, 643, 1327, 1663, 2371, 3643, 3847, 4327, 4951, 6091, 6571, 8263, 9151, 9187, 11239, 11383, 11863, 15307, 24007, 24151, 27847, 30091, 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

Offset: 1

Author

Emmanuel Vantieghem, Feb 05 2010

Keywords

• hard
• nonn

Comments

Primes 4*k+3 where half of the primitive roots are <= (p-1)/2.

The sequence is probably infinite.

Primes of the form 4m+1 always have as many primitive roots in [0,p/2] as in [p/2,p] (see A172480).

Crossrefs

Cf. A118818, A172480.

Programs

Maple

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
Nathaniel Johnston, Jun 26 2011

Mathematica

<< 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]

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

Extensions

More terms from Robert Israel, Nov 23 2019