OFFSET
1,1
COMMENTS
The identity (256*n-1)^2 - (256*n^2 - 2*n)*16^2 = 1 can be written as a(n)^2 - A158249(n)*16^2 = 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*(16^2*t-2)).
Vincenzo Librandi, X^2-AY^2=1
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
a(n) = 2*a(n-1) - a(n-2).
G.f.: x*(255 + x)/(1-x)^2.
E.g.f.: (-1 + 256*x)*exp(x). - G. C. Greubel, Apr 24 2022
MATHEMATICA
LinearRecurrence[{2, -1}, {255, 511}, 50]
PROG
(Magma) I:=[255, 511]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]];
(PARI) a(n) = 256*n - 1.
(SageMath) [256*n-1 for n in (1..50)] # G. C. Greubel, Apr 24 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 15 2009
STATUS
approved