A077958 Expansion of 1/(1-2*x^3).
1, 0, 0, 2, 0, 0, 4, 0, 0, 8, 0, 0, 16, 0, 0, 32, 0, 0, 64, 0, 0, 128, 0, 0, 256, 0, 0, 512, 0, 0, 1024, 0, 0, 2048, 0, 0, 4096, 0, 0, 8192, 0, 0, 16384, 0, 0, 32768, 0, 0, 65536, 0, 0, 131072, 0, 0, 262144, 0, 0, 524288, 0, 0, 1048576, 0, 0, 2097152, 0, 0, 4194304, 0, 0, 8388608, 0
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1002
- Darij Grinberg, Math 222: Enumerative Combinatorics, Fall 2019: Homework 1, Drexel University, Department of Mathematics, 2019.
- Index entries for linear recurrences with constant coefficients, signature (0,0,2).
Programs
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Magma
R
:=PowerSeriesRing(Integers(), 80); Coefficients(R!( 1/(1-2*x^3) )); // G. C. Greubel, Jun 23 2019 -
Mathematica
CoefficientList[Series[1/(1-2*x^3),{x,0,80}],x] (* Vladimir Joseph Stephan Orlovsky, Jan 30 2012 *) Riffle[Riffle[2^Range[0,30],0],0,{3,-1,3}] (* Harvey P. Dale, Dec 18 2012 *) -
PARI
Vec(1/(1-2*x^3)+O(x^80)) \\ Charles R Greathouse IV, Sep 27 2012
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PARI
a(n)= !(n%3)<<(n\3) \\ Ruud H.G. van Tol, Mar 28 2025
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Sage
(1/(1-2*x^3)).series(x, 80).coefficients(x, sparse=False) # G. C. Greubel, Jun 23 2019
Formula
From Stefano Spezia, Nov 26 2019: (Start)
a(n) = 2^(n/3) if 3 divides n, otherwise a(n) = 0 (see Exercise 2 in Grinberg).
E.g.f.: (1/3)*(exp(-(-2)^(1/3)*x) + exp(2^(1/3)*x) + exp((-1)^(2/3)*2^(1/3)*x)). (End)
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