|
%I #29 Sep 08 2022 08:45:08
%S 1,1,1,0,-2,-5,-8,-9,-5,7,28,54,73,64,1,-135,-335,-536,-602,-333,472,
%T 1879,3619,4887,4276,46,-9071,-22464,-35903,-40271,-22175,31824,
%U 126094,242539,327160,285687,1675,-609497,-1506356,-2404890,-2693927,-1476608,2145601,8461737,16254481,21901624
%N Expansion of (1-x)/(1-2*x+x^2+x^3).
%H G. C. Greubel, <a href="/A078001/b078001.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, -1).
%F a(n) = Sum_{k=0..floor(n/3)} (-1)^k*binomial(n-k, 2*k). - _Vladeta Jovovic_, Feb 10 2003
%F a(0)=1, a(n+1) = a(n) - Sum_{k=0..n-2} a(k). - _Alex Ratushnyak_, May 03 2012
%F a(0)=1, a(1)=1, a(2)=1, a(n) = 2*a(n-1)-a(n-2)-a(n-3). - _Harvey P. Dale_, Nov 03 2013
%t CoefficientList[Series[(1-x)/(1-2x+x^2+x^3),{x,0,50}],x] (* or *) LinearRecurrence[{2,-1,-1},{1,1,1},50] (* _Harvey P. Dale_, Nov 03 2013 *)
%o (Python)
%o a = [1]*1000
%o for n in range(55):
%o print(a[n], end=',')
%o sum=0
%o for k in range(n-1):
%o sum+=a[k]
%o a[n+1] = a[n]-sum
%o # from _Alex Ratushnyak_, May 03 2012
%o (PARI) Vec((1-x)/(1-2*x+x^2+x^3)+O(x^50)) \\ _Charles R Greathouse IV_, Sep 26 2012
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1-x)/( 1-2*x+x^2+x^3) )); // _G. C. Greubel_, Jun 27 2019
%o (Sage) ((1-x)/(1-2*x+x^2+x^3)).series(x, 50).coefficients(x, sparse=False) # _G. C. Greubel_, Jun 27 2019
%o (GAP) a:=[1,1,1];; for n in [4..50] do a[n]:=2*a[n-1]-a[n-2]-a[n-3]; od; a; # _G. C. Greubel_, Jun 27 2019
%Y Cf. A005251, A077856.
%K sign,easy
%O 0,5
%A _N. J. A. Sloane_, Nov 17 2002
|
