A151542 Generalized pentagonal numbers: a(n) = 12*n + 3*n*(n-1)/2.

0, 12, 27, 45, 66, 90, 117, 147, 180, 216, 255, 297, 342, 390, 441, 495, 552, 612, 675, 741, 810, 882, 957, 1035, 1116, 1200, 1287, 1377, 1470, 1566, 1665, 1767, 1872, 1980, 2091, 2205, 2322, 2442, 2565, 2691, 2820, 2952, 3087, 3225, 3366, 3510, 3657, 3807, 3960

Offset

0,2

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Leo Tavares, Illustration: Star Triangles

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

a(n) = a(n-1) + 3*n + 9 (with a(0)=0). - Vincenzo Librandi, Nov 26 2010

G.f.: 3*x*(4 - 3*x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 24 2011

From G. C. Greubel, May 26 2017: (Start)

a(n) = 3*n*(n+7)/2.

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

E.g.f.: (3/2)*(8*x + x^2)*exp(x). (End)

From Amiram Eldar, Feb 25 2022: (Start)

Sum_{n>=1} 1/a(n) = 121/490.

Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/21 - 319/4410. (End)

a(n) = A003154(n+1) - A060544(n). - Leo Tavares, Mar 26 2022

Mathematica

s=0; lst={}; Do[AppendTo[lst, s+=n], {n, 12, 6!, 3}]; lst (* Vladimir Joseph Stephan Orlovsky, Mar 05 2010 *)
LinearRecurrence[{3, -3, 1}, {0, 12, 27}, 50] (* or *) With[{nn = 50}, CoefficientList[Series[(3/2)*(8*x + x^2)*Exp[x], {x, 0, nn}], x] Range[0, nn]!] (* G. C. Greubel, May 26 2017 *)

Prog

(PARI) x='x+O('x^50); concat([0], Vec(serlaplace((3/2)*(8*x + x^2)*exp(x)))) \\ G. C. Greubel, May 26 2017
(PARI) a(n)=(3*n^2+21*n)/2 \\ Charles R Greathouse IV, Jun 16 2017

Crossrefs

The generalized pentagonal numbers b*n + 3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542.

Cf. A003154, A060544.

Sequence in context: A227854 A003346 A047723 * A069550 A079705 A068721

Adjacent sequences: A151539 A151540 A151541 * A151543 A151544 A151545

Keyword

easy,nonn

Author

N. J. A. Sloane, May 15 2009

Status

approved