OFFSET
1,5
COMMENTS
Also number of primitive polynomials of degree n over GF(2) whose second-highest coefficient is 0.
Always less than A011260 (and exactly one half of it when 2^n-1 is prime).
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 next-to-highest 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)=q-1 then
cnt := cnt+1;
fi;
od;
return cnt/n;
end;
for n in [1..32] do Print (a(n), ", "); od;
CROSSREFS
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