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a(n) = n*(2*n^2 + 5*n + 13)/2.
2

%I #14 Dec 17 2019 05:30:31

%S 0,10,31,69,130,220,345,511,724,990,1315,1705,2166,2704,3325,4035,

%T 4840,5746,6759,7885,9130,10500,12001,13639,15420,17350,19435,21681,

%U 24094,26680,29445,32395,35536,38874,42415,46165,50130,54316,58729,63375

%N a(n) = n*(2*n^2 + 5*n + 13)/2.

%H Vincenzo Librandi, <a href="/A163655/b163655.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F Row sums from A163652: a(n) = Sum_{m=1..n} (2*m*n + m + n + 6).

%F G.f.: x*(10 - 9*x + 5*x^2)/(x-1)^4.

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

%F E.g.f.: (1/2)*x*(20 + 11*x + 2*x^2)*exp(x). - _G. C. Greubel_, Aug 01 2017

%t CoefficientList[Series[x*(10-9*x+5*x^2)/(x-1)^4,{x,0,40}],x] (* _Vincenzo Librandi_, Mar 05 2012 *)

%t LinearRecurrence[{4,-6,4,-1}, {0,10,31,69}, 50] (* _G. C. Greubel_, Aug 01 2017 *)

%o (PARI) x='x+O('x^50); concat([0], Vec(x*(10-9*x+5*x^2)/(x-1)^4)) \\ _G. C. Greubel_, Aug 01 2017

%Y Cf. A163652.

%K nonn,easy

%O 0,2

%A _Vincenzo Librandi_, Aug 02 2009

%E Edited by _R. J. Mathar_, Aug 05 2009