A157913 a(n) = 64*n^2 - 16.

48, 240, 560, 1008, 1584, 2288, 3120, 4080, 5168, 6384, 7728, 9200, 10800, 12528, 14384, 16368, 18480, 20720, 23088, 25584, 28208, 30960, 33840, 36848, 39984, 43248, 46640, 50160, 53808, 57584, 61488, 65520, 69680, 73968, 78384, 82928, 87600, 92400, 97328, 102384

Offset

1,1

Comments

The identity (8*n^2 - 1)^2 - (64*n^2 - 16)*n^2 = 1 can be written as A157914(n)^2 - a(n)*n^2 = 1. - Vincenzo Librandi, Feb 09 2012

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

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

Formula

From Vincenzo Librandi, Feb 09 2012: (Start)

G.f.: -16*x*(3 + 6*x - x^2)/(x - 1)^3.

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

From Amiram Eldar, Mar 07 2023: (Start)

Sum_{n>=1} 1/a(n) = 1/32.

Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi-2)/64. (End)

Mathematica

LinearRecurrence[{3, -3, 1}, {48, 240, 560}, 50] (* Vincenzo Librandi, Feb 09 2012 *)
64*Range[40]^2-16 (* Harvey P. Dale, Jul 27 2012 *)

Prog

(Magma) I:=[48, 240, 560]; [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 09 2012
(PARI) for(n=1, 40, print1(64*n^2 - 16", ")); \\ Vincenzo Librandi, Feb 09 2012

Crossrefs

Cf. A157914.

Sequence in context: A211760 A333670 A230136 * A181773 A052683 A206054

Adjacent sequences: A157910 A157911 A157912 * A157914 A157915 A157916

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 09 2009

Status

approved