

A106290


Number of different orbit lengths of the 5step recursion mod n.


2



1, 3, 4, 4, 2, 9, 2, 6, 7, 6, 2, 11, 2, 6, 8, 8, 2, 9, 3, 8, 8, 6, 4, 12, 3, 6, 10, 8, 3, 18, 2, 10, 8, 6, 4, 11, 2, 6, 8, 12, 2, 18, 4, 8, 14, 9, 4, 16, 3, 9, 8, 8, 2, 12, 4, 12, 10, 6, 3, 22
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OFFSET

1,2


COMMENTS

Consider the 5step recursion x(k) = (x(k1)+x(k2)+x(k3)+x(k4)+x(k5)) 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 for each n. For n=8, there are 6 different lengths: 1, 2, 3, 6, 12 and 24. The maximum possible length of an orbit is A106303(n), the period of the Fibonacci 5step sequence mod n.


LINKS

Table of n, a(n) for n=1..60.
Eric Weisstein's World of Mathematics, Fibonacci nStep


PROG

(Python)
from itertools import count, product
def A106290(n):
bset, tset = set(), set()
for t in product(range(n), repeat=5):
t2 = t
for c in count(1):
t2 = t2[1:] + (sum(t2)%n, )
if t == t2:
bset.add(c)
tset.add(t)
break
if t2 in tset:
tset.add(t)
break
return len(bset) # Chai Wah Wu, Feb 22 2022


CROSSREFS

Cf. A106287 (orbits of 5step sequences), A106309 (primes that yield a simple orbit structure in 5step recursions).
Sequence in context: A209329 A353156 A081573 * A249618 A073498 A105736
Adjacent sequences: A106287 A106288 A106289 * A106291 A106292 A106293


KEYWORD

nonn,more


AUTHOR

T. D. Noe, May 02 2005


STATUS

approved



