OFFSET
1,1
COMMENTS
The identity (169*n+1)^2 - (169*n^2 + 2*n)*(13)^2 = 1 can be written as a(n)^2 - A158220(n)*(13)^2 = 1. - Vincenzo Librandi, Feb 02 2012
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*(13^2*t+2)).
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
G.f.: x*(170-x)/(1-x)^2. - Vincenzo Librandi, Feb 02 2012
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Feb 02 2012
MATHEMATICA
LinearRecurrence[{2, -1}, {170, 339}, 50] (* Vincenzo Librandi, Feb 02 2012 *)
PROG
(PARI) a(n)=169*n+1 \\ Charles R Greathouse IV, Dec 28 2011
(Magma) I:=[170, 339]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Feb 02 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 14 2009
STATUS
approved