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a(n) = floor(n*(n-1)*(n-2)/15).
2

%I #29 Oct 17 2024 08:25:39

%S 0,0,0,0,1,4,8,14,22,33,48,66,88,114,145,182,224,272,326,387,456,532,

%T 616,708,809,920,1040,1170,1310,1461,1624,1798,1984,2182,2393,2618,

%U 2856,3108,3374,3655,3952,4264,4592

%N a(n) = floor(n*(n-1)*(n-2)/15).

%H Vincenzo Librandi, <a href="/A011897/b011897.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,0,1,-3,3,-1).

%F From _R. J. Mathar_, Apr 15 2010:

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-5) - 3*a(n-6) + 3*a(n-7) - a(n-8).

%F G.f.: x^4*(1+x-x^2+x^3) / ( (1-x)^4*(1+x+x^2+x^3+x^4) ). (End)

%t CoefficientList[Series[x^4(1+x-x^2+x^3)/((1-x)^3*(1-x^5)),{x,0,45}],x] (* _Harvey P. Dale_, Feb 25 2011 *)

%t Floor[2*Binomial[Range[0, 50], 3]/5] (* _G. C. Greubel_, Oct 16 2024 *)

%o (Magma) [Floor(n*(n-1)*(n-2)/15): n in [0..50]]; // _Vincenzo Librandi_, Jul 07 2012

%o (SageMath) [2*binomial(n,3)//5 for n in range(51)] # _G. C. Greubel_, Oct 16 2024

%Y Cf. A011886.

%K nonn,easy

%O 0,6

%A _N. J. A. Sloane_