OFFSET
1,1
COMMENTS
The identity (8*n^2+1)^2 - (64*n^2+16)*n^2 = 1 can be written as A081585(n)^2 - a(n)*n^2 = 1. - Vincenzo Librandi, Feb 09 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Vincenzo Librandi, Feb 09 2012: (Start)
G.f.: x*(80+32*x+16*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
From Amiram Eldar, Mar 07 2023: (Start)
Sum_{n>=1} 1/a(n) = (coth(Pi/2)*Pi/2 - 1)/32.
Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/2)*Pi/2)/32. (End)
MAPLE
MATHEMATICA
64Range[50]^2+16 (* Harvey P. Dale, Mar 24 2011 *)
LinearRecurrence[{3, -3, 1}, {80, 272, 592}, 40] (* Vincenzo Librandi, Feb 09 2012 *)
PROG
(Magma) I:=[80, 272, 592]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 09 2012
(PARI) for(n=1, 40, print1(64*n^2 + 16", ")); \\ Vincenzo Librandi, Feb 09 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 09 2009
STATUS
approved