OFFSET
1,2
COMMENTS
Partial sums of A045991 (n^3-n^2). - Jeremy Gardiner, Jun 30 2013
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Project Euler, Sum square difference: Problem 6.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
a(n) = n*(n-1)*(n+1)*(3*n+2)/12.
G.f.: 2*x^2*(2+x)/(1-5*x+10*x^2-10*x^3+5*x^4-x^5). - Colin Barker, Jan 04 2012
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Apr 28 2012
a(n) = Sum_{k=1..n} (k-1)*k^2. - Michel Marcus, Nov 09 2012
From Amiram Eldar, Jan 10 2022: (Start)
Sum_{n>=2} 1/a(n) = 81*log(3)/5 - 9*sqrt(3)*Pi/5 - 192/25.
Sum_{n>=2} (-1)^n/a(n) = 18*sqrt(3)*Pi/5 - 48*log(2)/5 - 318/25. (End)
EXAMPLE
a(10) = 10 * 9 * 11 * 32 / 12 = 2640.
a(5) = 170 and the sum from 1 to 5 is 15, giving 1*(15-1)=14, 2*(15-2)=26, 2*(15-3)=36, 4*(15-4)=44 and 5*(15-5)=50; adding 14+26+36+44+50=170. Do the same for each n and get a(n). - J. M. Bergot, Oct 31 2014
MAPLE
a:=n->sum(j^3-j^2, j=0..n): seq(a(n), n=1..37); # Zerinvary Lajos, May 08 2008
MATHEMATICA
CoefficientList[Series[2*x*(2+x)/(1-5*x+10*x^2-10*x^3+ 5*x^4-x^5), {x, 0, 50}], x] (* Vincenzo Librandi, Apr 28 2012 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 4, 22, 70, 170}, 40] (* Harvey P. Dale, Jul 30 2019 *)
PROG
(Magma) I:=[0, 4, 22, 70, 170]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..45]]; // Vincenzo Librandi, Apr 28 2012
(PARI) a(n) = sum(k=1, n, (k-1)*k^2) \\ Michel Marcus, Nov 09 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ronald Arms (ron.arms(AT)stanfordalumni.org), Jan 23 2000
STATUS
approved