A158544 a(n) = 24*n^2 - 1.

23, 95, 215, 383, 599, 863, 1175, 1535, 1943, 2399, 2903, 3455, 4055, 4703, 5399, 6143, 6935, 7775, 8663, 9599, 10583, 11615, 12695, 13823, 14999, 16223, 17495, 18815, 20183, 21599, 23063, 24575, 26135, 27743, 29399, 31103, 32855, 34655, 36503, 38399, 40343

Offset

1,1

Comments

The identity (24*n^2 - 1)^2 - (144*n^2 - 12)*(2*n)^2 = 1 can be written as a(n)^2 - A158543(n)*A005843(n)^2 = 1.

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 14 2012: (Start)

G.f.: -x*(23 + 26*x - x^2)/(x - 1)^3.

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

From Amiram Eldar, Mar 06 2023: (Start)

Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(6)))*Pi/(2*sqrt(6)))/2.

Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(6)))*Pi/(2*sqrt(6)) - 1)/2. (End)

Mathematica

LinearRecurrence[{3, -3, 1}, {23, 95, 215}, 50] (* Vincenzo Librandi, Feb 14 2012 *)

Prog

(Magma) I:=[23, 95, 215]; [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 14 2012
(PARI) for(n=1, 40, print1(24*n^2 - 1", ")); \\ Vincenzo Librandi, Feb 14 2012

Crossrefs

Cf. A005843, A158543.

Sequence in context: A042032 A257976 A183011 * A154376 A155815 A231453

Adjacent sequences: A158541 A158542 A158543 * A158545 A158546 A158547

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 21 2009

Extensions

Comment rewritten by R. J. Mathar, Oct 16 2009

Status

approved