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a(n) = floor( n*(n-1)*(n-2)/11 ).
2

%I #28 Oct 10 2024 05:23:09

%S 0,0,0,0,2,5,10,19,30,45,65,90,120,156,198,248,305,370,445,528,621,

%T 725,840,966,1104,1254,1418,1595,1786,1993,2214,2451,2705,2976,3264,

%U 3570,3894,4238,4601,4984,5389,5814,6261,6731,7224,7740,8280,8844,9434,10049,10690

%N a(n) = floor( n*(n-1)*(n-2)/11 ).

%H G. C. Greubel, <a href="/A011893/b011893.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,0,0,0,0,0,0,0,1,-3,3,-1).

%F a(n) = +3*a(n-1) -3*a(n-2) +a(n-3) +a(n-11) -3*a(n-12) +3*a(n-13) -a(n-14). - _R. J. Mathar_, Apr 15 2010

%F G.f.: x^4*(2-x+x^2+2*x^3-2*x^4+2*x^5+x^6+x^9)/((1-x)^4*(1+x+x^2+x^3+x^4+x^5 +x^6+x^7+x^8+x^9+x^10)). - _Peter J. C. Moses_, Jun 02 2014

%t Table[Floor[n(n-1)(n-2)/11],{n,0,40}] (* or *)

%t LinearRecurrence[{3,-3,1,0,0,0,0,0,0,0,1,-3,3,-1}, {0,0,0,0,2,5,10,19,30,45, 65,90,120,156}, 50] (* _Harvey P. Dale_, Nov 23 2018 *)

%o (Magma) [Floor(6*Binomial(n,3)/11): n in [0..50]]; // _G. C. Greubel_, Oct 06 2024

%o (SageMath) [6*binomial(n,3)//11 for n in range(51)] # _G. C. Greubel_, Oct 06 2024

%Y Cf. A011886.

%K nonn,easy

%O 0,5

%A _N. J. A. Sloane_

%E a(41) onwards from _G. C. Greubel_, Oct 06 2024