>>11201209You can just compute the trace of X^n where X is the 3x3 diagonal with x,y,z on the diagonal.
Computing the characteristic polynomial of X gives X^3 - X^2 - X/2 - 1/6.
Solving X^3 - X^2 - X/2 - 1/6 = 0 will give x,y,z.
If you don't want to solve the cubic, use the recurrence A(n+3) = A(n+2) + A(n+1)/2 + A(n)/6.
This gives A(4) = A(3) + A(2)/2 + A(1)/6 = 3 + 1 + 1/6 = 25/6
(agreeing with
>>11201357)
A(5) = A(4) + A(3)/2 + A(2)/6 = (4 + 1/6) + (3/2) + (2/6) = 6
(agreeing with
>>11201689)
