OFFSET
1,1
COMMENTS
The identity (16*n^2+1)^2 - (64*n^2+8)*(2*n)^2 = 1 can be written as A108211(n)^2 - a(n)*A005843(n)^2 = 1. - rewritten by Bruno Berselli, Nov 16 2011
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
a(1)=72, a(2)=264, a(3)=584, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Nov 16 2011
G.f: x*(72 + 48*x + 8*x^2)/(1-x)^3. - Vincenzo Librandi, Feb 08 2012
From Amiram Eldar, Mar 05 2023: (Start)
Sum_{n>=1} 1/a(n) = (coth(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)) - 1)/16.
Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)))/16. (End)
MAPLE
MATHEMATICA
64Range[40]^2+8 (* or *) LinearRecurrence[{3, -3, 1}, {72, 264, 584}, 40] (* Harvey P. Dale, Nov 16 2011 *)
PROG
(Magma) I:=[72, 264, 584]; [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 08 2012
(PARI) for(n=1, 40, print1(64*n^2 + 8", ")); \\ Vincenzo Librandi, Feb 08 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 20 2009
STATUS
approved