A038518 Number of elements of GF(2^n) with trace 0 and subtrace 0.
0, 1, 1, 1, 6, 6, 16, 36, 56, 136, 256, 496, 1056, 2016, 4096, 8256, 16256, 32896, 65536, 130816, 262656, 523776, 1048576, 2098176, 4192256, 8390656, 16777216, 33550336, 67117056, 134209536, 268435456, 536887296, 1073709056, 2147516416
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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Maple
0,seq(1/4*2^k-1/4*(-1-I)^k-1/4*(-1+I)^k,k=1..40); seq(coeff(convert(series((-x^3+x^2+x)/((1-2*x)*(1+2*x+2*x^2)),x,50),polynom),x,i),i=0..40); # C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 16 2004
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Mathematica
LinearRecurrence[{0,2,4},{0,1,1,1},40] (* Harvey P. Dale, Mar 31 2020 *) -
PARI
concat(0, Vec(x*(1 + x - x^2) / ((1 - 2*x)*(1 + 2*x + 2*x^2)) + O(x^40))) \\ Colin Barker, Aug 02 2019
Formula
C(n, r+0)+C(n, r+4)+C(n, r+8)+... where r = 0 if n odd, r = 2 if n even.
G.f.: (-x^3+x^2+x)/[(1-2x)(1+2x+2x^2)].
a(0)=0; a(n) = ( 2^n - (-1-i)^n - (-1+i)^n )/4, i=sqrt(-1). - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 16 2004
a(n) = 2*a(n-2) + 4*a(n-3) for n>3. - Colin Barker, Aug 02 2019