OFFSET
1,1
COMMENTS
The identity (128*n+1)^2-(64*n^2+n)*16^2 = 1 can be written as a(n)^2-(A017066(n)+n)*16^2 = 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*(8^2*t+1)).
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
G.f.: x*(129-x)/(1-x)^2. - Vincenzo Librandi, Feb 10 2012
a(n) = 2*a(n-1)-a(n-2). - Vincenzo Librandi, Feb 10 2012
MAPLE
MATHEMATICA
128Range[50]+1 (* Harvey P. Dale, Mar 15 2011 *)
LinearRecurrence[{2, -1}, {129, 257}, 50] (* Vincenzo Librandi, Feb 10 2012 *)
PROG
(PARI) for(n=1, 50, print1(128*n + 1", ")); \\ Vincenzo Librandi, Feb 10 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 10 2009
STATUS
approved