OFFSET
1,1
COMMENTS
The identity (103680000*n^2 + 28800*n + 1)^2 - (3600*n^2 + n)*(1728000*n + 240)^2 = 1 can be written as A157863(n)^2 - a(n)*A157862(n)^2 = 1. - Vincenzo Librandi, Jan 25 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
G.f.: x*(3601 + 3599*x)/(1-x)^3. - Colin Barker, Jan 17 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jan 25 2012
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {3601, 14402, 32403}, 40] (* Vincenzo Librandi, Jan 25 2012 *)
PROG
(Magma) I:=[3601, 14402, 32403]; [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, Jan 25 2012
(PARI) for(n=1, 22, print1(3600*n^2 + n", ")); \\ Vincenzo Librandi, Jan 25 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 08 2009
STATUS
approved