A033486 a(n) = n*(n + 1)*(n + 2)*(n + 3)/2.

0, 12, 60, 180, 420, 840, 1512, 2520, 3960, 5940, 8580, 12012, 16380, 21840, 28560, 36720, 46512, 58140, 71820, 87780, 106260, 127512, 151800, 179400, 210600, 245700, 285012, 328860, 377580, 431520, 491040, 556512, 628320, 706860, 792540, 885780, 987012

Offset

0,2

Comments

a(n) is the area of an irregular quadrilateral with vertices at (1,1), (n+1, n+2), ((n+1)^2, (n+2)^2) and ((n+1)^3, (n+2)^3). - Art Baker, Dec 08 2018

Links

G. C. Greubel, Table of n, a(n) for n = 0..5000 (terms 0..680 from Vincenzo Librandi)

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = 6*A034827(n+3) = 12*A000332(n+3).

G.f.: 12*x/(1 - x)^5. - Colin Barker, Mar 01 2012

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) with a(0) = 0, a(1) = 12, a(2) = 60, a(3) = 180, a(4) = 420. - Harvey P. Dale, Feb 04 2015

E.g.f.: (24*x + 36*x^2 + 12*x^3 + x^4)*exp(x)/2. - Franck Maminirina Ramaharo, Dec 08 2018

From Amiram Eldar, Sep 04 2022: (Start)

Sum_{n>=1} 1/a(n) = 1/9.

Sum_{n>=1} (-1)^(n+1)/a(n) = 8*(3*log(2)-2)/9. (End)

Maple

[seq(12*binomial(n+3, 4), n=0..32)]; # Zerinvary Lajos, Nov 24 2006

Mathematica

Table[n*(n + 1)*(n + 2)*(n + 3)/2, {n, 0, 50}] (* David Nacin, Mar 01 2012 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 12, 60, 180, 420}, 40] (* Harvey P. Dale, Feb 04 2015 *)

Prog

(Magma) [n*(n+1)*(n+2)*(n+3)/2: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011
(PARI) a(n)=n*(n+1)*(n+2)*(n+3)/2 \\ Charles R Greathouse IV, Oct 07 2015
(GAP) List([0..40], n->n*(n+1)*(n+2)*(n+3)/2); # Muniru A Asiru, Dec 08 2018
(Sage) [12*binomial(n+3, 4) for n in range(40)] # G. C. Greubel, Dec 08 2018

Crossrefs

Cf. A000332, A050534, A033488, A033487, A060008.

Sequence in context: A332544 A279509 A008530 * A112415 A174642 A061624

Adjacent sequences: A033483 A033484 A033485 * A033487 A033488 A033489

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved