A078029 Expansion of (1-x)/(1-2*x^3).
1, -1, 0, 2, -2, 0, 4, -4, 0, 8, -8, 0, 16, -16, 0, 32, -32, 0, 64, -64, 0, 128, -128, 0, 256, -256, 0, 512, -512, 0, 1024, -1024, 0, 2048, -2048, 0, 4096, -4096, 0, 8192, -8192, 0, 16384, -16384, 0, 32768, -32768, 0, 65536, -65536, 0, 131072, -131072, 0, 262144, -262144, 0, 524288, -524288, 0
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (0,0,2).
Programs
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GAP
a:=[1,-1,0];; for n in [4..60] do a[n]:=2*a[n-3]; od; a; # G. C. Greubel, Aug 05 2019
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Magma
&cat[[2^n, -2^n, 0]: n in [0..60]]; // Vincenzo Librandi, May 10 2015
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Maple
seq(op([2^n,-2^n,0]), n=0..60); # Robert Israel, May 11 2015
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Mathematica
CoefficientList[Series[(1-x)/(1-2*x^3), {x, 0, 60}], x] (* Vincenzo Librandi, May 10 2015 *)
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PARI
my(x='x+O('x^60)); Vec((1-x)/(1-2*x^3)) \\ G. C. Greubel, Aug 05 2019
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Sage
((1-x)/(1-2*x^3)).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Aug 05 2019
Formula
G.f.: (1-x)/(1-2*x^3).
a(n) = (-1)^floor(4n/3)*(2-2*0^mod(n+1,3))^floor((n+1)/3). - Wesley Ivan Hurt, May 09 2015
a(n) = (4^(n/6)/6)*(2 - 2^(2/3) + 2^(5/3)*sin(Pi*(2*n/3 + 5/6)) - 4*sin(Pi*(2*n/3 + 3/2))). - Eric Simon Jacob, Jul 14 2024