A158557 a(n) = 225*n^2 + 15.

15, 240, 915, 2040, 3615, 5640, 8115, 11040, 14415, 18240, 22515, 27240, 32415, 38040, 44115, 50640, 57615, 65040, 72915, 81240, 90015, 99240, 108915, 119040, 129615, 140640, 152115, 164040, 176415, 189240, 202515, 216240, 230415, 245040, 260115, 275640, 291615

Offset

0,1

Comments

The identity (30*n^2 + 1)^2 - (225*n^2 + 15)*(2*n)^2 = 1 can be written as A158558(n)^2 - a(n)*A005843(n)^2 = 1.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

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

Formula

G.f.: 15*(1 + 13*x + 16*x^2)/(1-x)^3.

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

From Amiram Eldar, Mar 09 2023: (Start)

Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(15))*Pi/sqrt(15) + 1)/30.

Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(15))*Pi/sqrt(15) + 1)/30. (End)

Mathematica

LinearRecurrence[{3, -3, 1}, {15, 240, 915}, 50] (* Vincenzo Librandi, Feb 14 2012 *)
225*Range[0, 40]^2+15 (* Harvey P. Dale, Apr 06 2019 *)

Prog

(Magma) I:=[15, 240, 915]; [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=0, 22, print1(225*n^2 + 15", ")); \\ Vincenzo Librandi, Feb 14 2012

Crossrefs

Cf. A005843, A158558.

Sequence in context: A157456 A343527 A097262 * A220821 A090411 A154806

Adjacent sequences: A158554 A158555 A158556 * A158558 A158559 A158560

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 21 2009

Extensions

Comment rewritten, a(0) added by R. J. Mathar, Oct 16 2009

Status

approved