A154575 a(n) = 2*n^2 + 12*n + 4.

18, 36, 58, 84, 114, 148, 186, 228, 274, 324, 378, 436, 498, 564, 634, 708, 786, 868, 954, 1044, 1138, 1236, 1338, 1444, 1554, 1668, 1786, 1908, 2034, 2164, 2298, 2436, 2578, 2724, 2874, 3028, 3186, 3348, 3514, 3684, 3858, 4036, 4218, 4404, 4594, 4788

Offset

1,1

Comments

Sixth diagonal of A144562.

2*a(n) + 28 is a square.

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

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

Formula

From R. J. Mathar, Jan 05 2011: (Start)

a(n) = 2*A028881(n+3).

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

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Feb 26 2012

From Amiram Eldar, Feb 25 2023: (Start)

Sum_{n>=1} 1/a(n) = 1/28 - cot(sqrt(7)*Pi)*Pi/(4*sqrt(7)).

Sum_{n>=1} (-1)^(n+1)/a(n) = 31/84 - cosec(sqrt(7)*Pi)*Pi/(4*sqrt(7)). (End)

Mathematica

LinearRecurrence[{3, -3, 1}, {18, 36, 58}, 50] (* Vincenzo Librandi, Feb 26 2012 *)
Table[2n^2+12n+4, {n, 50}] (* Harvey P. Dale, Sep 18 2019 *)

Prog

(Magma) I:=[18, 36, 58]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 26 2012
(PARI) for(n=1, 50, print1(2*n^2+12*n+4", ")); \\ Vincenzo Librandi, Feb 26 2012

Crossrefs

Cf. A028881, A144562, A153238, A067076.

Sequence in context: A004961 A008600 A131766 * A344199 A097926 A087967

Adjacent sequences: A154572 A154573 A154574 * A154576 A154577 A154578

Keyword

nonn,easy

Author

Vincenzo Librandi, Jan 12 2009

Status

approved