|
|
A117188
|
|
Expansion of (1-x^2)/(1+x^2+x^4).
|
|
3
|
|
|
1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0, 1, 0, 1, 0, -2, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Periodic: repeat [1, 0, -2, 0, 1, 0].
Minton mentions that the subsequence a(2^i), i >= 1, oscillates between -2 and 1 (and does not converge 2-adically). - N. J. A. Sloane, Jul 09 2014
|
|
LINKS
|
|
|
FORMULA
|
G.f.: (1 - 2*x^2 + x^4)/(1-x^6).
a(n) = (1 + (-1)^n)/(-2 + 4^(floor((n-1)/3) - 2*floor((n-1)/6))). - Tani Akinari, Aug 02 2013
a(n) = a(n-6) for n>5.
a(n) = cos(n*Pi/2) * (cos(n*Pi/6) + sqrt(3)*sin(n*Pi/6)). (End)
E.g.f.: cos(sqrt(3)*x/2)*cosh(x/2) - sqrt(3)*sin(sqrt(3)*x/2)*sinh(x/2). - Ilya Gutkovskiy, Jun 27 2016
a(n) = cos((n+1)*Pi/3) - cos(2*(n+1)*Pi/3). - Ridouane Oudra, Dec 14 2021
|
|
MAPLE
|
|
|
MATHEMATICA
|
LinearRecurrence[{0, -1, 0, -1}, {1, 0, -2, 0}, 100] (* Harvey P. Dale, Jun 25 2022 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
sign,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|