A158487 a(n) = 64*n^2 - 8.

56, 248, 568, 1016, 1592, 2296, 3128, 4088, 5176, 6392, 7736, 9208, 10808, 12536, 14392, 16376, 18488, 20728, 23096, 25592, 28216, 30968, 33848, 36856, 39992, 43256, 46648, 50168, 53816, 57592, 61496, 65528, 69688, 73976, 78392, 82936, 87608, 92408, 97336, 102392

Offset

1,1

Comments

The identity (16*n^2 - 1)^2 - (64*n^2 - 8)*(2*n)^2 = 1 can be written as A141759(n)^2 - a(n)*A005843(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.: -8*x*(7 + 10*x - x^2)/(x - 1)^3.

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

From Amiram Eldar, Mar 05 2023: (Start)

Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)))/16.

Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)) - 1)/16. (End)

Mathematica

LinearRecurrence[{3, -3, 1}, {56, 248, 568}, 50] (* Vincenzo Librandi, Feb 09 2012 *)

Prog

(Magma) I:=[56, 248, 568]; [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 - 8", ")); \\ Vincenzo Librandi, Feb 09 2012

Crossrefs

Cf. A005843, A141759.

Sequence in context: A234762 A239597 A259039 * A212778 A205235 A205228

Adjacent sequences: A158484 A158485 A158486 * A158488 A158489 A158490

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 20 2009

Status

approved