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 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 (list; graph; refs; listen; history; text; internal format)
 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). LINKS 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 Cf. A192507 (GF(3^n)), A192508 (GF(5^n)), A192509 (GF(7^n)), A192510 (GF(11^n)), A192511 (GF(13^n)). Sequence in context: A287768 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

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Last modified October 17 16:42 EDT 2019. Contains 328120 sequences. (Running on oeis4.)