A157918 a(n) = 625*n^2 - 25.

600, 2475, 5600, 9975, 15600, 22475, 30600, 39975, 50600, 62475, 75600, 89975, 105600, 122475, 140600, 159975, 180600, 202475, 225600, 249975, 275600, 302475, 330600, 359975, 390600, 422475, 455600, 489975, 525600, 562475, 600600, 639975, 680600, 722475, 765600

Offset

1,1

Comments

The identity (50*n^2 - 1)^2 - (625*n^2 - 25)*(2*n)^2 = 1 can be written as A157919(n)^2 - a(n)*A005843(n)^2 = 1. - Vincenzo Librandi, Feb 10 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 10 2012: (Start)

G.f.: -25*x*(24 + 27*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 - cot(Pi/5)*Pi/5)/50.

Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/5)*Pi/5 - 1)/50. (End)

Mathematica

LinearRecurrence[{3, -3, 1}, {600, 2475, 5600}, 50] (* Vincenzo Librandi, Feb 10 2012 *)
625*Range[40]^2-25 (* Harvey P. Dale, May 19 2024 *)

Prog

(Magma) I:=[600, 2475, 5600]; [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 10 2012
(PARI) for(n=1, 40, print1(625*n^2 - 25", ")); \\ Vincenzo Librandi, Feb 10 2012

Crossrefs

Cf. A005843, A157919.

Sequence in context: A090222 A361900 A216058 * A092183 A048530 A223463

Adjacent sequences: A157915 A157916 A157917 * A157919 A157920 A157921

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 09 2009

Status

approved