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A152049
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Number of conjugacy classes of primitive elements in GF(2^n) which have trace 0
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6
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,5
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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).
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FORMULA
| a(n) = A192211(n)/n. [Joerg Arndt, Jul 03 2011]
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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.
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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;
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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: A010372 A199455 A197831 * A099887 A038220 A053151
Adjacent sequences: A152046 A152047 A152048 * A152050 A152051 A152052
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KEYWORD
| nonn,hard
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AUTHOR
| David A. Madore (david.madore(AT)ens.fr), Nov 21 2008
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EXTENSIONS
| More terms (13797...8911060) by Joerg Arndt, Jun 26 2011.
More terms (34636833...464635937) by Joerg Arndt, Jul 03 2011.
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