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-A158254(n)*(17)^2=1.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
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)).
Vincenzo Librandi, X^2-AY^2=1
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
G.f.: x*(290-x)/(1-x)^2. - Bruno Berselli, Mar 21 2011
a(n) = 2*a(n-1)-a(n-2).
MATHEMATICA
289Range[50]+1 (* Harvey P. Dale, Mar 21 2011 *)
LinearRecurrence[{2, -1}, {290, 579}, 50]
PROG
(Magma) I:=[290, 579]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]];
(PARI) a(n)=289*n+1 \\ Charles R Greathouse IV, Mar 22 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 15 2009
STATUS
approved
