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a(n) = (2*n^3 + 5*n^2 - 17*n)/2.
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%I #16 May 10 2021 11:36:45

%S -5,1,24,70,145,255,406,604,855,1165,1540,1986,2509,3115,3810,4600,

%T 5491,6489,7600,8830,10185,11671,13294,15060,16975,19045,21276,23674,

%U 26245,28995,31930,35056,38379,41905,45640,49590,53761,58159,62790,67660

%N a(n) = (2*n^3 + 5*n^2 - 17*n)/2.

%H Vincenzo Librandi, <a href="/A162259/b162259.txt">Table of n, a(n) for n = 1..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 A155551: a(n) = Sum_{m=1..n} (2*m*n + m + n - 9).

%F From _Vincenzo Librandi_, Mar 04 2012: (Start)

%F G.f.: x*(-5 + 21*x - 10*x^2)/(1-x)^4.

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

%t CoefficientList[Series[(-5+21*x-10*x^2)/(1-x)^4,{x,0,40}],x] (* or *) LinearRecurrence[{4, -6, 4, -1}, {-5, 1, 24, 70}, 50] (* _Vincenzo Librandi_, Mar 04 2012 *)

%t Table[(2n^3+5n^2-17n)/2,{n,40}] (* _Harvey P. Dale_, May 10 2021 *)

%Y Cf. A155551.

%K sign,easy

%O 1,1

%A _Vincenzo Librandi_, Jun 29 2009

%E New name from _Vincenzo Librandi_, Mar 04 2012