A038520 Number of elements of GF(2^n) with trace 1 and subtrace 0.
0, 1, 0, 3, 4, 6, 20, 28, 64, 136, 240, 528, 1024, 2016, 4160, 8128, 16384, 32896, 65280, 131328, 262144, 523776, 1049600, 2096128, 4194304, 8390656, 16773120, 33558528, 67108864, 134209536, 268451840, 536854528, 1073741824
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- F. Ruskey, Number of irreducible polynomials over GF(2) with given trace and subtrace
- F. Ruskey, Number of elements of GF(2^n) with given trace and subtrace
- Index entries for linear recurrences with constant coefficients, signature (0,2,4).
Programs
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Mathematica
LinearRecurrence[{0,2,4},{0,1,0,3},40] (* Harvey P. Dale, Oct 15 2017 *) -
PARI
concat(0, Vec(x*(1 + x^2) / ((1 - 2*x)*(1 + 2*x + 2*x^2)) + O(x^40))) \\ Colin Barker, Aug 02 2019
Formula
a(n) = C(n, r+0)+C(n, r+4)+C(n, r+8)+... where r = 1 if n odd, r = 3 if n even.
a(n) = 2*a(n-2) + 4*a(n-3), n > 3. - Paul Curtz, Feb 06 2008
From Colin Barker, Aug 02 2019: (Start)
G.f.: x*(1 + x^2) / ((1 - 2*x)*(1 + 2*x + 2*x^2)).
a(n) = (2^n + i*((-1-i)^n - (-1+i)^n)) / 4 for n>0, where i=sqrt(-1).
(End)