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A159833 a(n) = n^2*(n^2 + 15)/4.

%I A159833 #19 Sep 08 2022 08:45:44
%S A159833 0,4,19,54,124,250,459,784,1264,1944,2875,4114,5724,7774,10339,13500,
%T A159833 17344,21964,27459,33934,41500,50274,60379,71944,85104,100000,116779,
%U A159833 135594,156604,179974,205875,234484,265984,300564,338419,379750,424764
%N A159833 a(n) = n^2*(n^2 + 15)/4.
%H A159833 Vincenzo Librandi, <a href="/A159833/b159833.txt">Table of n, a(n) for n = 0..1000</a>
%H A159833 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A159833 a(n) = A008488(n+1)-2 = 4 - 15*A000292(n+1) + 6*A000332(n+4) + 20*A000217(n+1) - 15*(n+1).
%F A159833 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
%F A159833 G.f.: -x*(1+x)*(4*x^2-5*x+4)/(x-1)^5.
%F A159833 E.g.f.: x*(16 +22*x +6*x^2 +x^3)*exp(x)/4. - _G. C. Greubel_, May 19 2018
%p A159833 seq(n^2*(n^2+15)/4,n=0..80)
%t A159833 CoefficientList[Series[-x*(1 + x)*(4*x^2 - 5*x + 4)/(x-1)^5, {x, 0, 40}], x] (* _Vincenzo Librandi_, Dec 18 2012 *)
%t A159833 LinearRecurrence[{5,-10,10,-5,1},{0,4,19,54,124},40] (* _Harvey P. Dale_, May 30 2016 *)
%o A159833 (Magma) [n^2 * (n^2 + 15)/4: n in [0..40]]; // _Vincenzo Librandi_, Dec 18 2012
%o A159833 (PARI) for(n=0, 30, print1(n^2*(n^2 +15)/4, ", ")) \\ _G. C. Greubel_, May 19 2018
%Y A159833 Cf. A008488, A000292, A000332, A000217.
%K A159833 nonn,easy
%O A159833 0,2
%A A159833 _R. J. Mathar_, Apr 23 2009