A373946 Number of primitive polynomials of third degree over GF(m) with vanishing quadratic term with m = m(n) = A000961(n), for n >= 2.

1, 1, 0, 4, 3, 18, 8, 16, 18, 48, 48, 27, 80, 48, 108, 108, 72, 300, 144, 224, 180, 308, 192, 336, 560, 240, 648, 420, 576, 540, 648, 768, 1080, 1200, 912, 1360, 1008, 1352, 1188, 1584, 960, 2340, 1620, 4410, 2112, 2432, 1980, 2952, 1560, 2592, 2025, 4592, 2448, 4872, 4576

Offset: 2

Author

Martin Becker, Jun 23 2024

Keywords

• nonn

Comments

Apparently, a(n) = A373514(n) * A000010( 3 * A000961(n) - 3 ) * A025474(n) / 2, for n >= 2.

Examples

			For n=5, m=5, there are 20 primitive polynomials over GF(5) of the form x^3+a*x^2+b*x+c. Among these, 4 polynomials have a=0: x^3+3*x+2, x^3+3*x+3, x^3+4*x+2, and x^3+4*x+3. Thus, a(5) = 4.
		

Links

• Martin Becker, Table of n, a(n) for n = 2..400

Crossrefs

Cf. A000961, A319213, A373514.

Programs

PARI

is_max_o = (x1, x0, m, e)-> {
  for(i = 1, #e, if(x1^e[i] == x0, return(0))); x1^m == x0;
}
count_them = (q)-> {
  z = ffprimroot(ffgen(q, 'c));
  m = q^3 - 1; f = factor(m); d = #f~;
  e = vector(d, i, m/f[d + 1 - i, 1]);
  co = vector(q - 1, i, z^(i - 1));
  r = 0;
  for(a = 1, q - 1,
    for(b = 1, q - 1,
      p = co[1]*x^3 + co[a]*x + co[b];
      x1 = Mod(x, p); x0 = x1^0;
      if(is_max_o(x1, x0, m, e) && polisirreducible(p), r += 1)
    )
  );
  r;
}
print1(count_them(2));
for(q = 3, 64, if(isprimepower(q), print1(", ", count_them(q))))