OFFSET
1,1
COMMENTS
The identity (289*n+1)^2-(289*n^2+2*n)*(17)^2=1 can be written as A158255(n)^2-a(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 (3,-3,1).
FORMULA
G.f.: x*(291+287*x)/(1-x)^3. Also a(n) = (A158255(n)+1)*n. - Bruno Berselli, Mar 21 2011
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {291, 1160, 2607}, 50]
Table[289n^2+2n, {n, 40}] (* Harvey P. Dale, Jun 05 2013 *)
PROG
(Magma) I:=[291, 1160, 2607]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];
(PARI) a(n) = 289*n^2 + 2*n.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 15 2009
STATUS
approved