OFFSET
1,1
COMMENTS
The identity (29282*n^2 + 484*n + 1)^2 - (121*n^2 + 2*n)*(2662*n + 22)^2 = 1 can be written as a(n)^2 - A181679(n)*A157613(n)^2 = 1 (see also Bruno Berselli's comment at A181679). - Vincenzo Librandi, Feb 21 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Vincenzo Librandi, Feb 21 2012: (Start)
G.f: x*(-29767 - 28796*x - x^2)/(x-1)^3;
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {29767, 118097, 264991}, 50] (* Vincenzo Librandi, Feb 21 2012 *)
PROG
(Magma) I:=[29767, 118097, 264991]; [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(29282*n^2+484*n+1", ")); \\ Vincenzo Librandi, Feb 21 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 03 2009
STATUS
approved