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A038519 Number of elements of GF(2^n) with trace 0 and subtrace 1.

%I A038519 #36 Sep 08 2022 08:44:53
%S A038519 1,0,1,3,2,10,16,28,72,120,256,528,992,2080,4096,8128,16512,32640,
%T A038519 65536,131328,261632,524800,1048576,2096128,4196352,8386560,16777216,
%U A038519 33558528,67100672,134225920,268435456,536854528,1073774592
%N A038519 Number of elements of GF(2^n) with trace 0 and subtrace 1.
%H A038519 Colin Barker, <a href="/A038519/b038519.txt">Table of n, a(n) for n = 0..1000</a>
%H A038519 F. Ruskey, <a href="http://combos.org/TSpoly">Number of irreducible polynomials over GF(2) with given trace and subtrace</a>
%H A038519 F. Ruskey, <a href="http://combos.org/TSGF2">Number of elements of GF(2^n) with given trace and subtrace</a>
%H A038519 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,4).
%F A038519 a(n) = C(n, r+0) + C(n, r+4) + C(n, r+8) + ... where r = 2 if n odd, r = 0 if n even.
%F A038519 From _Colin Barker_, Aug 02 2019: (Start)
%F A038519 G.f.: (1 - x^2 - x^3) / ((1 - 2*x)*(1 + 2*x + 2*x^2)). - _Creighton Dement_, Apr 29 2005, corrected by _Colin Barker_, Aug 02 2019
%F A038519 a(n) = ((-1-i)^n + (-1+i)^n + 2^n) / 4 for n>0.
%F A038519 a(n) = 2*a(n-2) + 4*a(n-3) for n>3.
%F A038519 (End)
%o A038519 (PARI) Vec((1 - x^2 - x^3) / ((1 - 2*x)*(1 + 2*x + 2*x^2)) + O(x^40)) \\ _Colin Barker_, Aug 02 2019
%o A038519 (Magma) I:=[1,0,1,3]; [m le 4 select I[m] else 2*Self(m-2)+4*Self(m-3):m in [1..33]]; // _Marius A. Burtea_, Aug 02 2019
%Y A038519 Cf. A038503, A038505, A038518, A038520, A038521.
%K A038519 easy,nonn
%O A038519 0,4
%A A038519 _Frank Ruskey_