A192507 Number of conjugacy classes of primitive elements in GF(3^n) which have trace 0.

0, 0, 1, 2, 7, 14, 52, 104, 333, 870, 2571, 4590, 20440, 56736, 133782, 327558, 1265391, 2612694, 10188836, 20769420, 76562106

Offset: 1

Author

Joerg Arndt, Jul 03 2011

Keywords

• nonn
• hard
• more

Comments

Also number of primitive polynomials of degree n over GF(3) whose second-highest coefficient is 0.

Crossrefs

Cf. A152049 (GF(2^n)), A192507 (GF(5^n)), A192509 (GF(7^n)), A192510 (GF(11^n)), A192511 (GF(13^n)).

Cf. A027385 (number of primitive polynomials of degree n over GF(3)).

Programs

GAP

p := 3;
a := function(n)
    local q, k, cnt, x;
    q:=p^n;  k:=GF(p, n);  cnt:=0;
    for x in k do
        if Trace(k, GF(p), x)=0*Z(p) and Order(x)=q-1 then
            cnt := cnt+1;
        fi;
    od;
    return cnt/n;
end;
for n in [1..16] do  Print (a(n), ", ");  od;

Sage

# much more efficient
p=3; # choose characteristic
for n in range(1,66):
    F = GF(p^n, 'x')
    g = F.multiplicative_generator() # generator
    vt = vector(ZZ,p) # stats: trace
    m = p^n - 1 # size of multiplicative group
    # Compute all irreducible polynomials via Lyndon words:
    for w in LyndonWords(p,n): # digits of Lyndon words range form 1,..,p
        e = sum( (w[j]-1) * p^j for j in range(0,n) )
        if gcd(m, e) == 1: # primitive elements only
            f = g^e
            t = f.trace().lift(); # trace (over ZZ)
            vt[t] += 1
    print(vt[0]) # choose index 0,1,..,p-1 for different traces
# Joerg Arndt, Oct 03 2012

Formula

a(n) = A192212(n) / n.

Extensions

Added terms >=2571, Joerg Arndt, Oct 03 2012

a(18)-a(21) from Robin Visser, Apr 26 2024