A157915 a(n) = 625*n^2 + 25.

650, 2525, 5650, 10025, 15650, 22525, 30650, 40025, 50650, 62525, 75650, 90025, 105650, 122525, 140650, 160025, 180650, 202525, 225650, 250025, 275650, 302525, 330650, 360025, 390650, 422525, 455650, 490025, 525650, 562525, 600650, 640025, 680650, 722525, 765650

Offset

1,1

Comments

The identity (50*n^2 + 1)^2 - (625*n^2 + 25)*(2*n)^2 = 1 can be written as A157916(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: x*(650 + 575*x + 25*x^2)/(1-x)^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) = (coth(Pi/5)*Pi/5 - 1)/50.

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

Mathematica

625Range[40]^2+25  (* Harvey P. Dale, Apr 05 2011 *)
LinearRecurrence[{3, -3, 1}, {650, 2525, 5650}, 40] (* Vincenzo Librandi, Feb 10 2012 *)

Prog

(Magma) I:=[650, 2525, 5650]; [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 10 2012
(PARI) for(n=1, 40, print1(625*n^2 + 25", ")); \\ Vincenzo Librandi, Feb 10 2012

Crossrefs

Cf. A005843, A157916.

Sequence in context: A185666 A114047 A288142 * A158639 A162025 A251861

Adjacent sequences: A157912 A157913 A157914 * A157916 A157917 A157918

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 09 2009

Status

approved