A144449 a(n) = 4*(4 + 9*n^2 + 15*n).

16, 112, 280, 520, 832, 1216, 1672, 2200, 2800, 3472, 4216, 5032, 5920, 6880, 7912, 9016, 10192, 11440, 12760, 14152, 15616, 17152, 18760, 20440, 22192, 24016, 25912, 27880, 29920, 32032, 34216, 36472, 38800, 41200, 43672, 46216, 48832, 51520, 54280, 57112

Offset

0,1

Comments

A decimation: A061039(6n+5).

a(n) mod 9 = period 3: repeat 7,4,1 = A070403(n+1).

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

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

Formula

a(n) = a(n-1) + 24*(3*n+1) = a(n-1) + 72*n + 24, a(0)=16.

A061039(6n+2) = A061039(6n-4) + 24*(3n+1) = a(6n-4) + 72*n + 24, a(2)=16.

From G. C. Greubel, Mar 06 2022: (Start)

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

E.g.f.: 4*(4 + 24*x + 9*x^2)*exp(x). (End)

From Amiram Eldar, Mar 11 2022: (Start)

Sum_{n>=0} 1/a(n) = 1/12.

Sum_{n>=0} (-1)^n/a(n) = Pi/(18*sqrt(3)) + log(2)/18 - 1/12. (End)

Mathematica

Table[36n^2+60n+16, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {16, 112, 280}, 40] (* Harvey P. Dale, Apr 04 2020 *)

Prog

(Magma) [36*n^2 + 60*n + 16: n in [0..40]]; // Vincenzo Librandi, Aug 07 2011
(PARI) a(n)=36*n^2+60*n+16 \\ Charles R Greathouse IV, Jun 17 2017
(Sage) [(6*n+5)^2 - 9 for n in (0..40)] # G. C. Greubel, Mar 06 2022

Crossrefs

Cf. A061039, A070403.

Subsequence of A008590.

Sequence in context: A213754 A371114 A279425 * A035016 A000143 A258546

Adjacent sequences: A144446 A144447 A144448 * A144450 A144451 A144452

Keyword

nonn,easy

Author

Paul Curtz, Oct 06 2008

Extensions

Edited by Charles R Greathouse IV, Jul 25 2010

Status

approved