A144410 a(n) = 4*(3*n+1)*(3*n+2).

8, 80, 224, 440, 728, 1088, 1520, 2024, 2600, 3248, 3968, 4760, 5624, 6560, 7568, 8648, 9800, 11024, 12320, 13688, 15128, 16640, 18224, 19880, 21608, 23408, 25280, 27224, 29240, 31328, 33488, 35720, 38024, 40400, 42848, 45368, 47960, 50624, 53360, 56168, 59048, 62000, 65024, 68120, 71288, 74528, 77840, 81224, 84680, 88208, 91808, 95480

Offset

0,1

Comments

The sequence lists all numbers k such that k+1 is a square and k+4 is divisible by 12. - Bruno Berselli, Sep 28 2017

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

G.f.: 8*(1 + 7*x + x^2)/(1 - x)^3. - Michael De Vlieger, Sep 29 2017

a(n) = 8*A060544(n+1).

a(n) = A136016(2*n+1).

a(n) = a(m) + 36*(n - m)*(n + m + 1). For m = n-1, a(n) = a(n-1) + 72*n. - Bruno Berselli, Sep 29 2017

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n >= 3. - Klaus Purath, Jul 05 2020

E.g.f.: 4*(2 +18*x +9*x^2)*exp(x). - G. C. Greubel, Mar 27 2021

From Amiram Eldar, Dec 10 2022: (Start)

Sum_{n>=0} 1/a(n) = Pi/(12*sqrt(3)) (A244977).

Sum_{n>=0} (-1)^n/a(n) = log(2)/6. (End)

Maple

A144410:= n-> 4*(3*n+1)*(3*n+2); seq(A144410(n), n=0..60); # G. C. Greubel, Mar 27 2021

Mathematica

Table[4 (3 n + 1) (3 n + 2), {n, 0, 51}] (* or *)
CoefficientList[Series[8 (1 + 7 x + x^2)/(1 - x)^3, {x, 0, 51}], x] (* Michael De Vlieger, Sep 29 2017 *)

Prog

(Magma) [4*(3*n+1)*(3*n+2): n in [0..40]]; // Vincenzo Librandi, Aug 07 2011
(PARI) a(n)=4*(3*n+1)*(3*n+2) \\ Charles R Greathouse IV, Jun 17 2017
(Sage) [4*(3*n+1)*(3*n+2) for n in (0..60)] # G. C. Greubel, Mar 27 2021

Crossrefs

Cf. A060544, A136016, A244977.

Sequence in context: A204095 A061477 A069543 * A188149 A164755 A050799

Adjacent sequences: A144407 A144408 A144409 * A144411 A144412 A144413

Keyword

nonn,easy

Author

Paul Curtz, Sep 30 2008

Status

approved