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Consider the 3-step recursion x(k)=x(k-1)+x(k-2)+x(k-3) mod n. For any of the n^3 initial conditions x(1), x(2) and x(3) in Zn, the recursion has a finite period. Each of these n^3 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths (A106288) for each n. For instance, the orbits mod 8 have lengths of 1, 2, 4, 8, 16. Interestingly, for n=2^k and n=3^k, the number of orbits appear to be A039301 and A054879, respectively.

Consider the 5-step recursion x(k)=x(k-1)+x(k-2)+x(k-3)+x(k-4)+x(k-5) mod n. For any of the n^5 initial conditions x(1), x(2), x(3), x(4) and x(5) in Zn, the recursion has a finite period. Each of these n^5 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths (A106290). For instance, the 1468 orbits mod 8 have lengths of 1, 2, 3, 6, 12 and 24.

Consider the 4-step recursion x(k)=x(k-1)+x(k-2)+x(k-3)+x(k-4) mod n. For any of the n^4 initial conditions x(1), x(2), x(3) and x(4) in Zn, the recursion has a finite period. Each of these n^4 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths for each n. For n=8, there are 4 different lengths: 1, 5, 10 and 20. The maximum possible length of an orbit is the period of the Fibonacci 4-step sequence mod n, which is essentially A106295(n).

Consider the 4-step recursion x(k) = (x(k-1) + x(k-2) + x(k-3) + x(k-4)) mod n. For any of the n^4 initial conditions x(1), x(2), x(3) and x(4) in Zn, the recursion has a finite period. When n is a prime in this sequence, all of the orbits, except the one containing (0,0,0,0), have the same length.

For the prime 3 the orbit structure contains three orbits of length 1: (0,0,0,0), (1,1,1,1) and (2,2,2,2).