A106289 Number of different orbit lengths of the 4-step recursion mod n.

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

Offset

1,2

Comments

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 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 4-step 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 n-Step Number.

Crossrefs

Cf. A106286 (orbits of 4-step 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