OFFSET
1,1
COMMENTS
The identity (288*n + 1)^2 - (144*n^2 + n)*24^2 = 1 can be written as a(n)^2 - (A017522(n) + n)*24^2 = 1. - Vincenzo Librandi, Feb 10 2012
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 14 in the first table at p. 85, case d(t) = t*(12^2*t+1)).
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
G.f.: x*(289-x)/(1-x)^2. - Vincenzo Librandi, Feb 10 2012
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Feb 10 2012
MATHEMATICA
LinearRecurrence[{2, -1}, {289, 577}, 50] (* Vincenzo Librandi, Feb 10 2012 *)
PROG
(Magma) I:=[289, 577]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Feb 10 2012
(PARI) for(n=1, 50, print1(288*n + 1", ")); \\ Vincenzo Librandi, Feb 10 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 10 2009
STATUS
approved