OFFSET
1,1
COMMENTS
The identity (32*n^2 - 1)^2 - (256*n^2 - 16)*(2*n)^2 = 1 can be written as A158563(n)^2 - a(n)*A005843(n)^2 = 1. [rewritten by R. J. Mathar, Oct 16 2009]
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
G.f.: 16*x*(-15 - 18*x + x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 09 2023: (Start)
Sum_{n>=1} 1/a(n) = (4 - Pi)/128.
Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(2)*Pi - 4)/128. (End)
MATHEMATICA
16(16Range[40]^2-1) (* or *) LinearRecurrence[{3, -3, 1}, {240, 1008, 2288}, 40] (* Harvey P. Dale, Sep 13 2011 *)
PROG
(Magma) I:=[240, 1008, 2288]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 15 2012
(PARI) for(n=1, 50, print1(256*n^2-16", ")); \\ Vincenzo Librandi, Feb 15 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 21 2009
STATUS
approved