%I #14 Feb 16 2025 08:32:57
%S 1,4,6,28,3,24,10,220,91,12,130,240,343,40,168,1756,19,364,22,132,81,
%T 2068,26,1968,253,1372,2336,448,2557,672,16,14044,1143,76,108,4612,
%U 1411,88,3084,1860,11815,324,22,32092,13213,104,50,15792,2467,4012,168,17812
%N Number of orbits of the 4-step recursion mod n.
%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. Each of these n^4 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths (A106289) for each n. For instance, the 220 orbits mod 8 have lengths of 1, 5, 10 and 20.
%H D. D. Wall, <a href="http://www.jstor.org/stable/2309169">Fibonacci series modulo m</a>, Amer. Math. Monthly, 67 (1960), 525-532.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.
%Y Cf. A015134 (orbits of Fibonacci sequences), A106285 (orbits of 3-step sequences), A106287 (orbits of 5-step sequences), A106289 (number of different orbit lengths), A106308 (n producing a simple orbit structure).
%K nonn,changed
%O 1,2
%A _T. D. Noe_, May 02 2005