OFFSET
1,2
COMMENTS
Partial sums of A011379.
Antidiagonal sums of the convolution array A213819. - Clark Kimberling, Jul 04 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
a(n) = 2 * A001296(n-1) = (n-1)*n*(n+1)*(3*n-2)/12 (n>0). - Bruno Berselli, Apr 21 2010
a(n) = Sum_{i=1..n-1} binomial(i+1,i)*i^2. - Enrique Pérez Herrero, Jun 28 2014
G.f.: 2*x^2*(2*x+1) / (1 - x)^5. - Colin Barker, Jun 28 2014
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4. - Vincenzo Librandi, Jun 30 2014
a(n) = Sum_{k=1..n-1}k*((n-1)*n/2 + k) for n > 1. - J. M. Bergot, Feb 16 2018
From Amiram Eldar, Aug 23 2022: (Start)
Sum_{n>=2} 1/a(n) = 141/5 - 9*sqrt(3)*Pi/5 - 81*log(3)/5.
Sum_{n>=2} (-1)^n/a(n) = 18*sqrt(3)*Pi/5 + 48*log(2)/5 - 129/5. (End)
MATHEMATICA
a[n_]:=n^2; b[n_]:=n^3; c[n_]:=b[n]+a[n]; lst={}; s=0; Do[AppendTo[lst, s+=c[n]], {n, 0, 6!}]; lst
With[{r=Range[0, 50]}, Accumulate[r^2+r^3]] (* Harvey P. Dale, Jan 16 2011 *)
Rest[CoefficientList[Series[-2 x^2 * (2 x + 1)/(x - 1)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jun 30 2014 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 2, 14, 50, 130}, 25] (* G. C. Greubel, Sep 01 2016 *)
PROG
(PARI) concat(0, Vec(-2*x^2*(2*x+1)/(x-1)^5 + O(x^100))) \\ Colin Barker, Jun 28 2014
(PARI) a(n) = n*(n-1)*(n+1)*(3*n-2)/12 \\ Charles R Greathouse IV, Sep 01 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Joseph Stephan Orlovsky, Jan 03 2009
EXTENSIONS
Edited by Bruno Berselli, Jun 15 2010
Simpler definition as suggested by Wesley Ivan Hurt, Jun 29 2014
STATUS
approved