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 A157614(n)^2 - A181679(n)*a(n)^2 = 1 (see also Bruno Berselli's comment in 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 (2,-1).
FORMULA
From Vincenzo Librandi, Feb 21 2012: (Start)
G.f.: x*(2684 - 22*x)/(1-x)^2;
a(n) = 2*a(n-1) - a(n-2). (End)
MATHEMATICA
LinearRecurrence[{2, -1}, {2684, 5346}, 50] (* Vincenzo Librandi, Feb 21 2012 *)
2662 Range[40] + 22 (* Wesley Ivan Hurt, Nov 14 2023 *)
PROG
(Magma) I:=[2684, 5346]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Feb 21 2012
(PARI) for(n=1, 50, print1(2662*n + 22", ")); \\ Vincenzo Librandi, Feb 21 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 03 2009
STATUS
approved