OFFSET
1,1
COMMENTS
The identity (648*n^2 + 72*n + 1)^2 - (9*n^2 + n)*(216*n + 12)^2 = 1 can be written as a(n)^2 - A154517(n)*A154519(n)^2 = 1. This is the case s=3 of the identity (8*n^2*s^4 + 8*n*s^2 + 1)^2 - (n^2*s^2 + n)*(8*n*s^3 + 4*s)^2 = 1. - Vincenzo Librandi, Jan 30 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Colin Barker, Jan 25 2012: (Start)
G.f.: x*(721 + 574*x + x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=721, a(2)=2737, a(3)=6049. (End)
a(n) = 2*A161705(n)^2 - 1. - Bruno Berselli, Jan 31 2012
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {721, 2737, 6049}, 50] (* Vincenzo Librandi, Jan 30 2012 *)
PROG
(PARI) a(n)=648*n^2+72*n+1 \\ Charles R Greathouse IV, Dec 27 2011
(Magma) I:=[721, 2737, 6049]; [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 30 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Jan 11 2009
STATUS
approved