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a(n) = n * (2*n + 1) * (6*n^2 + 4*n + 1) / 3.
2

%I #15 Oct 21 2022 21:12:30

%S 0,11,110,469,1356,3135,6266,11305,18904,29811,44870,65021,91300,

%T 124839,166866,218705,281776,357595,447774,554021,678140,822031,

%U 987690,1177209,1392776,1636675,1911286,2219085,2562644,2944631,3367810,3835041

%N a(n) = n * (2*n + 1) * (6*n^2 + 4*n + 1) / 3.

%C Central terms of the triangle in A132121.

%H G. C. Greubel, <a href="/A132123/b132123.txt">Table of n, a(n) for n = 0..5000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5, -10, 10, -5, 1).

%F G.f.: x*(11 + 55*x + 29*x^2 + x^3)/(1-x)^5. - _Emeric Deutsch_, Aug 19 2007

%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=11, a(2)=110, a(3)=469, a(4)=1356. - _Harvey P. Dale_, Jun 02 2015

%F E.g.f.: x*(33 + 132*x + 86*x^2 + 12*x^3)*exp(x)/3. - _G. C. Greubel_, Mar 16 2019

%p seq((1/3)*n*(2*n+1)*(6*n^2+4*n+1),n=0..32); # _Emeric Deutsch_, Aug 19 2007

%t Table[n(2n+1)(6n^2+4n+1)/3,{n,0,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,11,110,469,1356},40] (* _Harvey P. Dale_, Jun 02 2015 *)

%o (PARI) {a(n) = n*(2*n+1)*(6*n^2+4*n+1)/3}; \\ _G. C. Greubel_, Mar 16 2019

%o (Magma) [n*(2*n+1)*(6*n^2+4*n+1)/3: n in [0..40]]; // _G. C. Greubel_, Mar 16 2019

%o (Sage) [n*(2*n+1)*(6*n^2+4*n+1)/3 for n in (0..40)] # _G. C. Greubel_, Mar 16 2019

%K nonn,easy

%O 0,2

%A _Reinhard Zumkeller_, Aug 12 2007