A172118 a(n) = n*(n+1)*(5*n^2 - n - 3)/2.

0, 1, 45, 234, 730, 1755, 3591, 6580, 11124, 17685, 26785, 39006, 54990, 75439, 101115, 132840, 171496, 218025, 273429, 338770, 415170, 503811, 605935, 722844, 855900, 1006525, 1176201, 1366470, 1578934, 1815255, 2077155, 2366416, 2684880

Offset

0,3

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

From Bruno Berselli, May 07 2010: (Start)

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

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

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)

G.f. x*(1+40*x+19*x^2)/(1-x)^5. - R. J. Mathar, Nov 17 2011

From G. C. Greubel, Apr 15 2022: (Start)

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

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

Mathematica

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 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 45, 234, 730}, 40] (* Harvey P. Dale, Jul 20 2022 *)

Prog

(Magma) [n*(n+1)*(5*n^2-n-3)/2: n in [0..50]]; // Vincenzo Librandi, Aug 20 2014
(SageMath) [n*(n+1)*(5*n^2-n-3)/2 for n in (0..50)] # G. C. Greubel, Apr 15 2022

Crossrefs

Cf. A172117.

Sequence in context: A251444 A169717 A246420 * A127073 A089549 A178540

Adjacent sequences: A172115 A172116 A172117 * A172119 A172120 A172121

Keyword

nonn,easy

Author

Vincenzo Librandi, Jan 26 2010

Extensions

Formula simplified and sequence A172117 corrected by Bruno Berselli, May 07 2010

Status

approved