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A011849 a(n) = floor(binomial(n,3)/3).

%I A011849 #33 Jun 13 2023 11:16:07
%S A011849 0,0,0,0,1,3,6,11,18,28,40,55,73,95,121,151,186,226,272,323,380,443,
%T A011849 513,590,674,766,866,975,1092,1218,1353,1498,1653,1818,1994,2181,2380,
%U A011849 2590,2812,3046,3293,3553,3826
%N A011849 a(n) = floor(binomial(n,3)/3).
%H A011849 Vincenzo Librandi, <a href="/A011849/b011849.txt">Table of n, a(n) for n = 0..1000</a>
%H A011849 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (4, -6, 3, 3, -6, 3, 3, -6, 4, -1).
%F A011849 G.f.: x^4*(1-x+x^2)*(1-x^2+x^3)/((1-x)^3*(1-x^9)). - _Ralf Stephan_, Mar 05 2004
%F A011849 From _R. J. Mathar_, Apr 15 2010: (Start)
%F A011849 a(n) = 4*a(n-1) - 6*a(n-2) + 3*a(n-3) + 3*a(n-4) - 6*a(n-5) + 3*a(n-6) + 3*a(n-7) - 6*a(n-8) + 4*a(n-9) - a(n-10).
%F A011849 G.f.: x^4*(x^2-x+1)*(x^3-x^2+1) / ( (-1+x)^4*(x^6+x^3+1) ). (End)
%F A011849 a(n) = (1/54) * ( n^3 - 3*n - 6 + [6,8,4,-12,8,4,-30,8,4](mod 9) ). - _Ralf Stephan_, Aug 11 2013
%p A011849 seq(floor(binomial(n,3)/3), n=0..42); # _Zerinvary Lajos_, Jan 12 2009
%t A011849 CoefficientList[Series[x^4*(x^2-x+1)*(x^3-x^2+1)/((-1+x)^4*(x^6+x^3+1)),{x,0,50}],x] (* _Vincenzo Librandi_, Jun 19 2012 *)
%o A011849 (Magma) [Floor(Binomial(n,3)/3): n in [0..50]]; // _Vincenzo Librandi_, Jun 19 2012
%Y A011849 A column of triangle A011857.
%K A011849 nonn,easy
%O A011849 0,6
%A A011849 _N. J. A. Sloane_