OFFSET
1,1
COMMENTS
The identity (128*n^2 - 32*n + 1)^2 - (4*n^2 - n)*(64*n - 8)^2 = 1 can be written as A157331(n)^2 - A033991(n)*a(n)^2 = 1. This is the case s=2 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 29 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, Jan 29 2012: (Start)
a(n) = 2*a(n-1) - a(n-2).
G.f.: x*(8*x+56)/(x-1)^2. (End)
a(n) = 8*A004771(n-1). - Michel Marcus, Aug 19 2018
MATHEMATICA
LinearRecurrence[{2, -1}, {56, 120}, 50] (* Vincenzo Librandi, Jan 29 2012 *)
PROG
(Magma) I:=[56, 120]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Jan 29 2012
(PARI) for(n=1, 40, print1(64*n - 8", ")); \\ Vincenzo Librandi, Jan 29 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 27 2009
STATUS
approved