OFFSET
0,1
COMMENTS
Let A_{9,3} = [0,0,0,1; 0,0,1,1; 0,1,1,1; 1,1,1,1], a unit-primitive matrix (see [Jeffery]). Then a(n) = Trace([A_{9,3}]^n).
LINKS
L. E. Jeffery, Unit-primitive matrices
Index entries for linear recurrences with constant coefficients, signature (2, 3, -1, -1).
FORMULA
G.f.: (4-6*x-6*x^2+x^3)/((1+x)*(1-3*x+x^3)).
a(n) = 2*a(n-1)+3*a(n-2)-a(n-3)-a(n-4), {a(m)}={4,2,10,23}, m=0,1,2,3.
a(n) = Sum_{k=1..4} ((x_k)^3-2*(x_k))^n, x_k=2*(-1)^(k-1)*cos(k*Pi/9).
a(n) = (-1)^n+(1+2*cos(Pi/9))^n+(1-cos(Pi/9)+sqrt(3)*sin(Pi/9))^n + (1-cos(Pi/9)-sqrt(3)*sin(Pi/9))^n. - L. Edson Jeffery, Dec 15 2011
a(n) = (-1)^n + 3*A147704(n). - R. J. Mathar, Oct 08 2016
MATHEMATICA
CoefficientList[Series[(4-6x-6x^2+x^3)/((1+x)(1-3x+x^3)), {x, 0, 30}], x] (* or *) LinearRecurrence[{2, 3, -1, -1}, {4, 2, 10, 23}, 30] (* Harvey P. Dale, Apr 22 2011 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
L. Edson Jeffery, Apr 05 2011
STATUS
approved