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a(n) = n*(n+1)*(5*n^2 - n - 3)/2.
1

%I #29 Jul 20 2022 15:14:37

%S 0,1,45,234,730,1755,3591,6580,11124,17685,26785,39006,54990,75439,

%T 101115,132840,171496,218025,273429,338770,415170,503811,605935,

%U 722844,855900,1006525,1176201,1366470,1578934,1815255,2077155,2366416,2684880

%N a(n) = n*(n+1)*(5*n^2 - n - 3)/2.

%H Vincenzo Librandi, <a href="/A172118/b172118.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F From _Bruno Berselli_, May 07 2010: (Start)

%F a(n) = n*(n*(n+1)*(20*n-17)/6) - Sum_{i=0..n-1} ( i*(i+1)*(20*i-17)/2 ).

%F a(n) = n*(n+1)*(5*n^2-n-3)/2.

%F More generally: n*(n*(n+1)*(2*d*n-2*d+3)/6) - Sum_{i=0..n-1} ( i*(i+1)*(2*d*i-2*d+3)/6, i=0..n-1 ) = n*(n+1)*(3*d*n^2 - d*n + 4*n - 2*d + 2)/12; in this sequence is d=10. (End)

%F G.f. x*(1+40*x+19*x^2)/(1-x)^5. - _R. J. Mathar_, Nov 17 2011

%F From _G. C. Greubel_, Apr 15 2022: (Start)

%F a(n) = 12*binomial(n+3,4) - 78*binomial(n+2,3) + 19*binomial(n+1,2).

%F E.g.f.: (1/2)*x*(2 + 43*x + 34*x^2 + 5*x^3)*exp(x). (End)

%t Table[n*(n+1)*(5*n^2-n-3)/2, {n, 0, 40}] (* or *) CoefficientList[Series[x(1 +40x +19x^2)/(1-x)^5, {x, 0, 40}], x] (* _Vincenzo Librandi_, Aug 20 2014 *)

%t LinearRecurrence[{5,-10,10,-5,1},{0,1,45,234,730},40] (* _Harvey P. Dale_, Jul 20 2022 *)

%o (Magma) [n*(n+1)*(5*n^2-n-3)/2: n in [0..50]]; // _Vincenzo Librandi_, Aug 20 2014

%o (SageMath) [n*(n+1)*(5*n^2-n-3)/2 for n in (0..50)] # _G. C. Greubel_, Apr 15 2022

%Y Cf. A172117.

%K nonn,easy

%O 0,3

%A _Vincenzo Librandi_, Jan 26 2010

%E Formula simplified and sequence A172117 corrected by _Bruno Berselli_, May 07 2010