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Number of orbits of the 5-step recursion mod n.
4

%I #13 Mar 24 2024 07:56:45

%S 1,8,5,96,5,56,7,1468,203,40,11,1312,13,56,25,23400,17,6392,193,480,

%T 35,88,555,37180,2505,104,15539,672,293,280,151,374292,55,136,35,

%U 199744,37,6128,65,7340,41,392,1899,1056,1015,6648,313775,627280,14413,20040,85

%N Number of orbits 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 (A106290). For instance, the 1468 orbits mod 8 have lengths of 1, 2, 3, 6, 12 and 24.

%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="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.

%Y Cf. A015134 (orbits of Fibonacci sequences), A106285 (orbits of 3-step sequences), A106286 (orbits of 4-step sequences), A106290 (number of different orbit lengths), A106309 (n producing a simple orbit structure).

%K nonn

%O 1,2

%A _T. D. Noe_, May 02 2005