OFFSET
1,1
COMMENTS
The identity (2048*n^2+128*n+1)^2-(16*n^2+n)*(512*n+16)^2=1 can be written as a(n)^2-A157474(n)*A157475(n)^2=1. [rewritten by Bruno Berselli, Aug 22 2011]
This is the case s=4 of the identity (8*n^2*s^4+8*n*s^2+1)^2 - (n^2*s^2+n)*(8*n*s^3+4*s)^2 = 1. - Bruno Berselli, Jan 25 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Harvey P. Dale, Aug 15 2011: (Start)
G.f.: x*(-x^2-1918*x-2177)/(x-1)^3.
a(1)=2177, a(2)=8449, a(3)=18817, a(n)=3*a(n-1)-3*a(n-2)+a(n-3). (End)
MATHEMATICA
Table[2048n^2+128n+1, {n, 30}] (* or *) LinearRecurrence[{3, -3, 1}, {2177, 8449, 18817}, 30] (* Harvey P. Dale, Aug 15 2011 *)
PROG
(PARI) a(n)=2048*n^2+128*n+1 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 01 2009
STATUS
approved