OFFSET
1,1
COMMENTS
The identity (18*n^2 + 1)^2 - (81*n^2 + 9)*(2*n)^2 = 1 can be written as A157889(n)^2 - a(n)*A005843(n+1)^2 = 1. - Vincenzo Librandi, Feb 05 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 05 2012: (Start)
G.f: x*(90 + 63*x + 9*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/3)*Pi/3 - 1)/18.
Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/3)*Pi/3)/18. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: 9*(exp(x)*(9*x^2 + + 9*x + 1) - 1).
a(n) = 9*A247792(n). (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {90, 333, 738}, 40] (* Vincenzo Librandi, Feb 05 2012 *)
81*Range[40]^2+9 (* Harvey P. Dale, Aug 05 2015 *)
PROG
(Magma) I:=[90, 333, 738]; [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 05 2012
(PARI) for(n=1, 40, print1(81*n^2 + 9", ")); \\ Vincenzo Librandi, Feb 05 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 08 2009
STATUS
approved