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%I #22 Mar 24 2024 04:03:24
%S 1,3,4,4,2,9,2,6,7,6,2,11,2,6,8,8,2,9,3,8,8,6,4,12,3,6,10,8,3,18,2,10,
%T 8,6,4,11,2,6,8,12,2,18,4,8,14,9,4,16,3,9,8,8,2,12,4,12,10,6,3,22
%N Number of different orbit lengths of the 5-step recursion mod n.
%C 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 for each n. For n=8, there are 6 different lengths: 1, 2, 3, 6, 12 and 24. The maximum possible length of an orbit is A106303(n), the period of the Fibonacci 5-step sequence mod n.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.
%o (Python)
%o from itertools import count,product
%o def A106290(n):
%o bset, tset = set(), set()
%o for t in product(range(n),repeat=5):
%o t2 = t
%o for c in count(1):
%o t2 = t2[1:] + (sum(t2)%n,)
%o if t == t2:
%o bset.add(c)
%o tset.add(t)
%o break
%o if t2 in tset:
%o tset.add(t)
%o break
%o return len(bset) # _Chai Wah Wu_, Feb 22 2022
%Y Cf. A106287 (orbits of 5-step sequences), A106309 (primes that yield a simple orbit structure in 5-step recursions).
%K nonn,more
%O 1,2
%A _T. D. Noe_, May 02 2005
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