OFFSET
1,1
COMMENTS
The identity (256*n+1)^2-(256*n^2+2*n)*(16)^2=1 can be written as A158231(n)^2-a(n)*(16)^2=1.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1
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)).
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.: 2*x*(129+127*x)/(1-x)^3.
E.g.f.: 2*exp(x)*x*(129 + 128*x). - Stefano Spezia, Aug 19 2025
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {258, 1028, 2310}, 50]
PROG
(Magma) I:=[258, 1028, 2310]; [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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 14 2009
STATUS
approved