

A106288


Number of different orbit lengths of the 3step recursion mod n.


1



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, 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, 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
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OFFSET

1,2


COMMENTS

Consider the 3step recursion x(k)=x(k1)+x(k2)+x(k3) 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 3step sequence mod n.


LINKS

Table of n, a(n) for n=1..95.
Eric Weisstein's World of Mathematics, Fibonacci nStep


CROSSREFS

Cf. A106285 (orbits of 3step sequences), A106307 (primes that yield a simple orbit structure in 3step recursions).
Sequence in context: A304182 A256284 A322979 * A013633 A016559 A104566
Adjacent sequences: A106285 A106286 A106287 * A106289 A106290 A106291


KEYWORD

nonn


AUTHOR

T. D. Noe, May 02 2005


STATUS

approved



