A158488 a(n) = 64*n^2 + 8.

72, 264, 584, 1032, 1608, 2312, 3144, 4104, 5192, 6408, 7752, 9224, 10824, 12552, 14408, 16392, 18504, 20744, 23112, 25608, 28232, 30984, 33864, 36872, 40008, 43272, 46664, 50184, 53832, 57608, 61512, 65544, 69704, 73992, 78408, 82952, 87624, 92424, 97352, 102408

Offset

1,1

Comments

The identity (16*n^2+1)^2 - (64*n^2+8)*(2*n)^2 = 1 can be written as A108211(n)^2 - a(n)*A005843(n)^2 = 1. - rewritten by Bruno Berselli, Nov 16 2011

Links

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

Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]

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

Formula

a(1)=72, a(2)=264, a(3)=584, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Nov 16 2011

G.f: x*(72 + 48*x + 8*x^2)/(1-x)^3. - Vincenzo Librandi, Feb 08 2012

From Amiram Eldar, Mar 05 2023: (Start)

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

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

Maple

A158488:=n->64*n^2+8: seq(A158488(n), n=1..50); # Wesley Ivan Hurt, Apr 08 2017

Mathematica

64Range[40]^2+8 (* or *) LinearRecurrence[{3, -3, 1}, {72, 264, 584}, 40] (* Harvey P. Dale, Nov 16 2011 *)

Prog

(Magma) I:=[72, 264, 584]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 08 2012
(PARI) for(n=1, 40, print1(64*n^2 + 8", ")); \\ Vincenzo Librandi, Feb 08 2012

Crossrefs

Cf. A005843, A108211.

Sequence in context: A250749 A019507 A303081 * A165139 A305222 A316800

Adjacent sequences: A158485 A158486 A158487 * A158489 A158490 A158491

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 20 2009

Status

approved