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

%I #33 May 08 2022 10:08:55

%S 2,11,33,74,140,237,371,548,774,1055,1397,1806,2288,2849,3495,4232,

%T 5066,6003,7049,8210,9492,10901,12443,14124,15950,17927,20061,22358,

%U 24824,27465,30287,33296,36498,39899,43505,47322,51356,55613,60099,64820

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

%H Vincenzo Librandi, <a href="/A085786/b085786.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 a(n) = A000217(n) + n^3.

%F From _Colin Barker_, Jan 20 2014: (Start)

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

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

%F E.g.f.: (x/2)*(4 + 7*x + 2*x^2)*exp(x). - _G. C. Greubel_, Aug 24 2017

%t CoefficientList[Series[(x + 1) (x + 2) / (x - 1)^4, {x, 0, 40}], x] (* or *) LinearRecurrence[{4, -6, 4, -1}, {2, 11, 33, 74}, 40] (* _Vincenzo Librandi_, Aug 14 2017 *)

%o (PARI) t(n)=n*(n+1)/2;

%o vector(40, i, t(i)+i^3)

%o (Magma) [n*(2*n^2 + n + 1)/2: n in [1..40]]; // _Vincenzo Librandi_, Aug 14 2017

%Y Cf. A000217 [t(n)], A000096 [t(n)+n], A005449 [t(n)+n^2].

%Y a(n) = A110449(n, n).

%K nonn,easy

%O 1,1

%A _Jon Perry_, Jul 23 2003

%E Name changed by _Wesley Ivan Hurt_, Apr 30 2022