OFFSET
1,1
COMMENTS
The identity (22*n^2 - 1)^2 - (121*n^2 - 11)*(2*n)^2 = 1 can be written as A158540(n)^2 - a(n)*A005843(n)^2 = 1. - Vincenzo Librandi, Feb 21 2012
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
From Vincenzo Librandi, Feb 21 2012: (Start)
G.f.: 11*x*(-10 - 13*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/sqrt(11))*Pi/sqrt(11))/22.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(11))*Pi/sqrt(11) - 1)/22. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {110, 473, 1078}, 50] (* Vincenzo Librandi, Feb 21 2012 *)
121 Range[40]^2-11 (* or *) CoefficientList[Series[(11(x^2-13x-10))/(x-1)^3, {x, 0, 40}], x] (* Harvey P. Dale, Aug 16 2021 *)
PROG
(Magma) I:=[110, 473, 1078]; [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 21 2012
(PARI) for(n=1, 40, print1(121*n^2 - 11", ")); \\ Vincenzo Librandi, Feb 21 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 21 2009
EXTENSIONS
Edited by N. J. A. Sloane, Oct 12 2009
STATUS
approved