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

%I #11 Sep 08 2022 08:46:24

%S 1,15,90,340,975,2331,4900,9360,16605,27775,44286,67860,100555,144795,

%T 203400,279616,377145,500175,653410,842100,1072071,1349755,1682220,

%U 2077200,2543125,3089151,3725190,4461940,5310915,6284475,7395856,8659200,10089585

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

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F From _Vincenzo Librandi_, Nov 13 2019: (Start)

%F a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).

%F G.f.: x*(1+9*x+15*x^2+5*x^3)/(1-x)^6. (End)

%t CoefficientList[Series[(1+9x+15x^2+5x^3)/(1-x)^6,{x,0,33}],x] (* _Vincenzo Librandi_, Nov 13 2019 *)

%o (Magma) [(n^2+n^3+n^4+n^5)/4: n in [1..40]] // _Vincenzo Librandi_, Nov 13 2019

%Y Cf. A329126, A329000.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_, Nov 12 2019