OFFSET
1,1
COMMENTS
The identity (8*n^2 - 1)^2 - (64*n^2 - 16)*n^2 = 1 can be written as A157914(n)^2 - a(n)*n^2 = 1. - Vincenzo Librandi, Feb 09 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Vincenzo Librandi, Feb 09 2012: (Start)
G.f.: -16*x*(3 + 6*x - x^2)/(x - 1)^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) = 1/32.
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi-2)/64. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {48, 240, 560}, 50] (* Vincenzo Librandi, Feb 09 2012 *)
64*Range[40]^2-16 (* Harvey P. Dale, Jul 27 2012 *)
PROG
(Magma) I:=[48, 240, 560]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // 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