

A152049


Number of conjugacy classes of primitive elements in GF(2^n) which have trace 0.


6



0, 0, 1, 1, 3, 2, 9, 9, 23, 29, 89, 72, 315, 375, 899, 1031, 3855, 3886, 13797, 12000, 42328, 59989, 178529, 138256, 647969, 859841, 2101143, 2370917, 9204061, 8911060, 34636833, 33556537, 105508927, 168423669, 464635937
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OFFSET

1,5


COMMENTS

Also number of primitive polynomials of degree n over GF(2) whose secondhighest coefficient is 0.
Always less than A011260 (and exactly one half of it when 2^n1 is prime).


LINKS

Table of n, a(n) for n=1..35.


FORMULA

a(n) = A192211(n)/n. [Joerg Arndt, Jul 03 2011]


EXAMPLE

a(3)=1 because of the two primitive degree 3 polynomials over GF(2), namely t^3+t+1 and t^3+t^2+1, only the former has a zero nexttohighest coefficient.
Similarly, a(13)=315, because of half (4096) of the 8192 elements of GF(2^13) have trace 0 and all except 0 (since 1 has trace 1) are primitive, so there are 4095/13=315 conjugacy classes of primitive elements of trace 0.


PROG

(GAP)
a := function(n)
local q, k, cnt, x; q:=2^n; k:=GF(2, n); cnt:=0;
for x in k do
if Trace(k, GF(2), x)=0*Z(2) and Order(x)=q1 then
cnt := cnt+1;
fi;
od;
return cnt/n;
end;
for n in [1..32] do Print (a(n), ", "); od;


CROSSREFS

Cf. A192507 (GF(3^n)), A192508 (GF(5^n)), A192509 (GF(7^n)), A192510 (GF(11^n)), A192511 (GF(13^n)).
Sequence in context: A199455 A197831 A244995 * A246788 A099887 A274827
Adjacent sequences: A152046 A152047 A152048 * A152050 A152051 A152052


KEYWORD

nonn,hard,more


AUTHOR

David A. Madore, Nov 21 2008


EXTENSIONS

More terms (13797...8911060) by Joerg Arndt, Jun 26 2011.
More terms (34636833...464635937) by Joerg Arndt, Jul 03 2011.


STATUS

approved



