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 a(n)^2 - A033991(n)*A157330(n)^2 = 1 (see also the second part of the comment at A157330). - 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 (3,-3,1).
FORMULA
From Vincenzo Librandi, Jan 29 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(-97 - 158*x - x^2)/(x-1)^3. (End)
E.g.f.: exp(x)*(128*x^2 + 96*x + 1) - 1. - Stefano Spezia, May 02 2024
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {97, 449, 1057}, 40] (* Vincenzo Librandi, Jan 29 2012 *)
PROG
(Magma) I:=[97, 449, 1057]; [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 29 2012
(PARI) for(n=1, 40, print1(128*n^2 - 32*n + 1", ")); \\ Vincenzo Librandi, Jan 29 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 27 2009
STATUS
approved