A027480 a(n) = n*(n+1)*(n+2)/2.
0, 3, 12, 30, 60, 105, 168, 252, 360, 495, 660, 858, 1092, 1365, 1680, 2040, 2448, 2907, 3420, 3990, 4620, 5313, 6072, 6900, 7800, 8775, 9828, 10962, 12180, 13485, 14880, 16368, 17952, 19635, 21420, 23310, 25308, 27417, 29640, 31980, 34440
Offset: 0
Examples
Row sums of n consecutive integers, starting at 0, seen as a triangle: . 0 | 0 3 | 1 2 12 | 3 4 5 30 | 6 7 8 9 60 | 10 11 12 13 14 105 | 15 16 17 18 19 20
Links
- T. D. Noe, Table of n, a(n) for n = 0..1000
- Douglas N. Arnold, Richard S. Falk, and Ragnar Winther, Finite element exterior calculus, homological techniques, and applications, Acta Numerica 15 (2006), 1-155.
- Steve Butler and Pavel Karasik, A note on nested sums, J. Int. Seq. 13 (2010), 10.4.4.
- Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 9.
- Solomon Gartenhaus, Odd Order Pandiagonal Latin and Magic Cubes in Three and Four Dimensions, arXiv:math/0210275 [math.CO], 2002.
- Index entries for sequences related to dominoes
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
Programs
-
Magma
[n*(n+1)*(n+2)/2: n in [0..40]]; // Vincenzo Librandi, Nov 14 2014
-
Maple
[seq(3*binomial(n+2,3),n=0..37)]; # Zerinvary Lajos, Nov 24 2006 a := n -> add((j+n)*(n+2)/3,j=0..n): seq(a(n),n=0..35); # Zerinvary Lajos, Dec 17 2006
-
Mathematica
Table[(m^3 - m)/2, {m, 36}] (* Zerinvary Lajos, Mar 21 2007 *) LinearRecurrence[{4,-6,4,-1},{0,3,12,30},40] (* Harvey P. Dale, Oct 10 2012 *) CoefficientList[Series[3 x / (x - 1)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 14 2014 *) With[{nn=50},Total/@TakeList[Range[0,(nn(nn+1))/2-1],Range[nn]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Jun 02 2019 *)
-
PARI
a(n)=3*binomial(n+2,3) \\ Charles R Greathouse IV, May 23 2011
-
Python
def a(n): return (n**3+3*n**2+2*n)//2 # _Torlach Rush, Jun 16 2024
Formula
a(n) = n*binomial(2+n, 2). - Zerinvary Lajos, Jan 10 2006
a(n) = A007531(n+2)/2. - Zerinvary Lajos, Jul 17 2006
Starting with offset 1 = binomial transform of [3, 9, 9, 3, 0, 0, 0]. - Gary W. Adamson, Oct 25 2007
From R. J. Mathar, Apr 07 2009: (Start)
G.f.: 3*x/(x-1)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
a(n) = Sum_{i=0..n} n*(n - i) + 2*i. - Bruno Berselli, Jan 13 2016
From Ilya Gutkovskiy, Aug 07 2016: (Start)
E.g.f.: x*(6 + 6*x + x^2)*exp(x)/2.
a(n) = Sum_{k=0..n} A045943(k).
Sum_{n>=1} 1/a(n) = 1/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (8*log(2) - 5)/2 = 0.2725887222397812... = A016639/10. (End)
a(n-1) = binomial(n^2,2)/n for n > 0. - Jonathan Sondow, Jan 07 2018
For k > 1, Sum_{i=0..n^2-1} (k+i)^2 = (k*n + a(k-1))^2 + A126275(k). - Charlie Marion, Apr 23 2021
Comments