OFFSET
1,1
COMMENTS
The identity (256*n - 1)^2 - (256*n^2 - 2*n)*16^2 = 1 can be written as A158250(n)^2 - a(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 (3,-3,1).
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(254 + 258*x)/(1-x)^3.
E.g.f.: 2*x*(127 + 128*x)*exp(x). - G. C. Greubel, Apr 24 2022
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {254, 1020, 2298}, 50]
PROG
(Magma) I:=[254, 1020, 2298]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];
(PARI) a(n) = 256*n^2-2*n.
(SageMath) [2*n*(128*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