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A192507
Number of conjugacy classes of primitive elements in GF(3^n) which have trace 0.
7
0, 0, 1, 2, 7, 14, 52, 104, 333, 870, 2571, 4590, 20440, 56736, 133782, 327558, 1265391, 2612694, 10188836, 20769420, 76562106
OFFSET
1,4
COMMENTS
Also number of primitive polynomials of degree n over GF(3) whose second-highest coefficient is 0.
FORMULA
a(n) = A192212(n) / n.
PROG
(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
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)).
Sequence in context: A018281 A018592 A123752 * A018622 A018668 A173756
KEYWORD
nonn,hard,more
AUTHOR
Joerg Arndt, Jul 03 2011
EXTENSIONS
Added terms >=2571, Joerg Arndt, Oct 03 2012
a(18)-a(21) from Robin Visser, Apr 26 2024
STATUS
approved