A099721 a(n) = n^2*(2*n+1).

0, 3, 20, 63, 144, 275, 468, 735, 1088, 1539, 2100, 2783, 3600, 4563, 5684, 6975, 8448, 10115, 11988, 14079, 16400, 18963, 21780, 24863, 28224, 31875, 35828, 40095, 44688, 49619, 54900, 60543, 66560, 72963, 79764, 86975, 94608, 102675, 111188, 120159, 129600

Offset: 0

Author

Douglas Winston (douglas.winston(AT)srupc.com), Nov 07 2004

Keywords

• nonn
• easy

Comments

For a right triangle with sides of lengths 8*n^3 + 12*n^2 + 8*n + 2, 4*n^4 + 8*n^3 + 4*n^2, and 4*n^4 + 8*n^3 + 12*n^2 + 8*n + 2, dividing the area by the perimeter gives a(n). - J. M. Bergot, Jul 30 2013

This sequence is the difference between the centered icosahedral (or cuboctahedral) numbers (A005902(n)) and the centered octagonal pyramidal numbers (A000447(n+1)). - Peter M. Chema, Jan 09 2016

a(n) is the sum of the integers in the closed interval (n-1)*n to n*(n+1). - J. M. Bergot, Apr 19 2017

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Row n=3 of A229079.

Cf. A000578, A001093, A011379, A015237, A027444, A033431, A033562, A034262, A053698, A061317, A066023, A071568, A098547.

Cf. A000290, A000447, A002378, A005408, A005902, A024196.

Programs

Magma

[n^2*(2*n+1): n in [0..50]]; // Vincenzo Librandi, May 01 2011

Maple

A099721 := proc(n) n^2*(2*n+1) ; end proc:
seq(A099721(n),n=0..10) ;

Mathematica

a[n_]:=2*n^3+n^2; (* Vladimir Joseph Stephan Orlovsky, Dec 21 2008 *)
LinearRecurrence[{4,-6,4,-1},{0,3,20,63},40] (* Harvey P. Dale, Aug 19 2022 *)

PARI

a(n) = ceil(sum(i=n^2-(n-1), n^2+(n-1), if(!issquare(4*i+1), (2*i+1+sqrt(4*i+1))/2, 0))); \\ Michel Marcus, Nov 14 2014, after Richard R. Forberg

Formula

G.f.: x*(3 + 8*x + x^2)/(x-1)^4.

a(n) = A024196(n) - A024196(n-1). - Philippe Deléham, May 07 2012

a(n) = ceiling(Sum_{i=n^2-(n-1)..n^2+(n-1)} s(i)), for n > 0 and integer i, where s(i) are the real solutions to x = i + sqrt(x), and the summation range excludes the integer solutions which occur where i is an oblong number (A002378). The fractional portion of the summation converges to 2/3 for large n. If s(i) is replaced with i, then the summation equals n^2*(2*n-1) = A015237. - Richard R. Forberg, Oct 15 2014

a(n) = A005902(n) - A000447(n+1). - Peter M. Chema, Jan 09 2016

From Amiram Eldar, May 17 2022: (Start)

Sum_{n>=1} 1/a(n) = Pi^2/6 + 4*log(2) - 4.

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/12 - Pi - 2*log(2) + 4. (End)

From Elmo R. Oliveira, Aug 08 2025: (Start)

E.g.f.: x*(1 + 2*x)*(3 + x)*exp(x).

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

a(n) = A000290(n)*A005408(n). (End)