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%I #15 Sep 24 2013 20:55:26
%S 0,0,1,1,3,2,9,9,23,29,89,72,315,375,899,1031,3855,3886,13797,12000,
%T 42328,59989,178529,138256,647969,859841,2101143,2370917,9204061,
%U 8911060,34636833,33556537,105508927,168423669,464635937
%N Number of conjugacy classes of primitive elements in GF(2^n) which have trace 0.
%C Also number of primitive polynomials of degree n over GF(2) whose second-highest coefficient is 0.
%C Always less than A011260 (and exactly one half of it when 2^n-1 is prime).
%F a(n) = A192211(n)/n. [_Joerg Arndt_, Jul 03 2011]
%e 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 next-to-highest coefficient.
%e 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.
%o (GAP)
%o a := function(n)
%o local q,k,cnt,x; q:=2^n; k:=GF(2,n); cnt:=0;
%o for x in k do
%o if Trace(k, GF(2), x)=0*Z(2) and Order(x)=q-1 then
%o cnt := cnt+1;
%o fi;
%o od;
%o return cnt/n;
%o end;
%o for n in [1..32] do Print (a(n), ", "); od;
%Y Cf. A192507 (GF(3^n)), A192508 (GF(5^n)), A192509 (GF(7^n)), A192510 (GF(11^n)), A192511 (GF(13^n)).
%K nonn,hard,more
%O 1,5
%A _David A. Madore_, Nov 21 2008
%E More terms (13797...8911060) by _Joerg Arndt_, Jun 26 2011.
%E More terms (34636833...464635937) by _Joerg Arndt_, Jul 03 2011.
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