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A011886 a(n) = floor(n*(n-1)*(n-2)/4). 1

%I #25 Jun 13 2023 10:54:03

%S 0,0,0,1,6,15,30,52,84,126,180,247,330,429,546,682,840,1020,1224,1453,

%T 1710,1995,2310,2656,3036,3450,3900,4387,4914,5481,6090,6742,7440,

%U 8184,8976,9817,10710,11655,12654,13708,14820,15990,17220,18511,19866,21285

%N a(n) = floor(n*(n-1)*(n-2)/4).

%H Vincenzo Librandi, <a href="/A011886/b011886.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 1, 1, -3, 3, -1).

%F a(n) = +3*a(n-1) -3*a(n-2) +a(n-3) +a(n-4) -3*a(n-5) +3*a(n-6) -a(n-7). G.f.: x^3*(1+3*x+2*x^3) / ( (-1+x)^4*(1+x)*(x^2+1) ). - _R. J. Mathar_, Apr 15 2010

%F f(n) = floor(Sum_{k=0..n} n*(k+1)/2) for n >= -2. - _William A. Tedeschi_, Sep 10 2010

%t Table[Floor[(n(n-1)(n-2))/4],{n,0,50}] (* or *) LinearRecurrence[ {3,-3,1,1,-3,3,-1},{0,0,0,1,6,15,30},50] (* _Harvey P. Dale_, Feb 25 2012 *)

%t CoefficientList[Series[x^3*(1+3*x+2*x^3)/((-1+x)^4*(1+x)*(x^2+1)),{x,0,50}],x] (* _Vincenzo Librandi_, Jul 07 2012 *)

%o (Magma) [Floor(n*(n-1)*(n-2)/4): n in [0..50]]; // _Vincenzo Librandi_, Jul 07 2012

%K nonn,easy

%O 0,5

%A _N. J. A. Sloane_.

%E More terms from _William A. Tedeschi_, Sep 10 2010

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