

A106289


Number of different orbit lengths of the 4step recursion mod n.


1



1, 2, 2, 3, 2, 4, 4, 4, 4, 4, 3, 5, 3, 8, 3, 5, 3, 8, 3, 5, 7, 4, 4, 7, 3, 6, 6, 9, 4, 6, 2, 6, 6, 6, 6, 10, 5, 6, 6, 6, 5, 14, 2, 6, 5, 8, 3, 9, 7, 4, 6, 7, 2, 12, 5, 12, 6, 7, 4, 7, 3, 4, 8, 7, 5, 8, 4, 7, 7, 12, 3, 14, 4, 10, 4, 8, 10, 12, 2, 7, 8, 6, 2, 15, 6, 3, 8, 8, 2, 10, 8, 9, 3, 6, 6, 11, 2, 14, 8
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OFFSET

1,2


COMMENTS

Consider the 4step recursion x(k)=x(k1)+x(k2)+x(k3)+x(k4) 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 for each n. For n=8, there are 4 different lengths: 1, 5, 10 and 20. The maximum possible length of an orbit is the period of the Fibonacci 4step sequence mod n, which is essentially A106295(n).


LINKS

Table of n, a(n) for n=1..99.
Eric Weisstein's World of Mathematics, Fibonacci nStep


CROSSREFS

Cf. A106286 (orbits of 4step sequences).
Sequence in context: A007897 A180783 A290731 * A332436 A165418 A307797
Adjacent sequences: A106286 A106287 A106288 * A106290 A106291 A106292


KEYWORD

nonn


AUTHOR

T. D. Noe, May 02 2005


STATUS

approved



