A168559 a(n) = n^2 + a(n-1), with a(1)=0.

0, 4, 13, 29, 54, 90, 139, 203, 284, 384, 505, 649, 818, 1014, 1239, 1495, 1784, 2108, 2469, 2869, 3310, 3794, 4323, 4899, 5524, 6200, 6929, 7713, 8554, 9454, 10415, 11439, 12528, 13684, 14909, 16205, 17574, 19018, 20539, 22139, 23820, 25584

Offset

1,2

Comments

Sum of the first n perfect squares (A000330), minus 1.

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Formula

a(n) = n^3/3 + n^2/2 + n/6 - 1. - Gary Detlefs, Jun 30 2010

For n>1, a(n) = 2^2 + 3^2 + ... + n^2. - Washington Bomfim, Feb 15 2011

G.f.: x*(4-3*x+x^2)/(1-x)^4. - Colin Barker, Feb 03 2012

a(n) = A000330(n) - 1. - Reinhard Zumkeller, Feb 03 2012

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) with a(1)=0, a(2)=4, a(3)=13, a(4)=29. - Harvey P. Dale, Dec 07 2013

Mathematica

k=-1; lst={}; Do[k=n^2+k; AppendTo[lst, k], {n, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 05 2009 *)
RecurrenceTable[{a[1]==0, a[n]==n^2+a[n-1]}, a, {n, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 4, 13, 29}, 50] (* Harvey P. Dale, Dec 07 2013 *)
Accumulate[Range[50]^2] - 1 (* Paolo Xausa, Oct 04 2024 *)

Prog

(Haskell)
a168559 n = a168559_list !! (n-1)
a168559_list = scanl (+) 0 $ drop 2 a000290_list
-- Reinhard Zumkeller, Feb 03 2012
(PARI) a(n)=n^3/3+n^2/2+n/6-1 \\ Charles R Greathouse IV, Oct 16 2015

Crossrefs

Cf. A000330, A000290.

Sequence in context: A340002 A368570 A135039 * A212247 A213801 A301886

Adjacent sequences: A168556 A168557 A168558 * A168560 A168561 A168562

Keyword

nonn,easy

Author

Vincenzo Librandi, Nov 30 2009

Status

approved