OFFSET
1,1
COMMENTS
The identity (10368*n^2 + 288*n + 1)^2 - (36*n^2 + n)*(1728*n + 24)^2 = 1 can be written as a(n)^2 - A157324(n)*A157325(n)^2 = 1 (see also second part of the comment at A157324). - Vincenzo Librandi, Jan 26 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
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jan 26 2012
G.f.: x*(-x^2 - 10078*x - 10657)/(x-1)^3. - Vincenzo Librandi, Jan 26 2012
a(n) = 2*A017533(6n)^2 - 1. - Bruno Berselli, Jan 29 2012
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {10657, 42049, 94177}, 50] (* Vincenzo Librandi, Jan 26 2012 *)
PROG
(Magma) I:=[10657, 42049, 94177]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Jan 26 2012
(PARI) for(n=1, 22, print1(10368*n^2 + 288*n + 1", ")); \\ Vincenzo Librandi, Jan 26 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 27 2009
STATUS
approved