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%I #11 Mar 24 2024 04:24:14
%S 2,3,5,31,43,53,79,83,89,97,109,131,137,139,151,199,229,233,239,257,
%T 283,313,317,359,367,389,433,443,479,487,569,571,577,601,617,641,643,
%U 659,673,677,769
%N Primes that yield a simple orbit structure in 4-step recursions.
%C 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.
%C 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).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.
%Y Cf. A106286 (orbits of 4-step sequences).
%K nonn
%O 1,1
%A _T. D. Noe_, May 02 2005, revised May 12 2005
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