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%I #11 Mar 24 2024 07:57:01
%S 1,3,2,4,2,6,3,5,3,6,4,8,3,6,4,6,3,9,3,8,6,8,2,10,3,5,4,8,3,12,2,7,8,
%T 5,6,12,2,6,6,10,3,12,3,11,6,6,3,12,5,9,6,7,3,12,8,9,6,6,2,16,3,6,7,8,
%U 6,16,2,6,4,12,2,15,3,6,6,8,10,10,3,12,5,5,3,16,6,7,6,14,2,18,6,8,4,6,6
%N Number of different orbit lengths of the 3-step recursion mod n.
%C 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 for each n. For n=8, there are 5 different lengths: 1, 2, 4, 8 and 16. The maximum possible length of an orbit is A046738(n), the period of the Fibonacci 3-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>.
%Y Cf. A106285 (orbits of 3-step sequences), A106307 (primes that yield a simple orbit structure in 3-step recursions).
%K nonn
%O 1,2
%A _T. D. Noe_, May 02 2005
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