OFFSET
1,1
COMMENTS
The identity (50*n^2 - 1)^2 - (625*n^2 - 25)*(2*n)^2 = 1 can be written as a(n)^2 - A157918(n)*A005843(n)^2 = 1. - Vincenzo Librandi, Feb 10 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 10 2012: (Start)
G.f.: -x*(49 + 52*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 - cot(Pi/(5*sqrt(2)))*Pi/(5*sqrt(2)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(5*sqrt(2)))*Pi/(5*sqrt(2)) - 1)/2. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {49, 199, 449}, 50] (* Vincenzo Librandi, Feb 10 2012 *)
50 Range[40]^2-1 (* Harvey P. Dale, Dec 15 2018 *)
PROG
(Magma) I:=[49, 199, 449]; [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 10 2012
(PARI) for(n=1, 40, print1(50*n^2 - 1", ")); \\ Vincenzo Librandi, Feb 10 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 09 2009
STATUS
approved