OFFSET
1,1
COMMENTS
The identity (289*n-1)^2-(289*n^2-2*n)*(17)^2=1 can be written as a(n)^2-A158252(n)*(17)^2=1.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1
E. J. Barbeau, Polynomial Excursions, Chapter 10: Diophantine equations (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(17^2*t-2)).
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
a(n) = 2*a(n-1)-a(n-2).
G.f.: x*(288+x)/(1-x)^2.
MATHEMATICA
289Range[50]-1 (* Harvey P. Dale, Nov 01 2011 *)
LinearRecurrence[{2, -1}, {288, 577}, 50]
PROG
(Magma) I:=[288, 577]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]];
(PARI) a(n) = 289*n - 1.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 15 2009
STATUS
approved