login
A157951
a(n) = 128*n + 1.
1
129, 257, 385, 513, 641, 769, 897, 1025, 1153, 1281, 1409, 1537, 1665, 1793, 1921, 2049, 2177, 2305, 2433, 2561, 2689, 2817, 2945, 3073, 3201, 3329, 3457, 3585, 3713, 3841, 3969, 4097, 4225, 4353, 4481, 4609, 4737, 4865, 4993, 5121, 5249, 5377, 5505
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
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)).
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
A157951:=n->128*n + 1: seq(A157951(n), n=1..80); # Wesley Ivan Hurt, Jan 24 2017
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
Cf. A017066.
Sequence in context: A034072 A178228 A235879 * A043383 A250814 A249679
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 10 2009
STATUS
approved