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A106290 Number of different orbit lengths of the 5-step 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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 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 5-step sequence mod n.

LINKS

Table of n, a(n) for n=1..60.

Eric Weisstein's World of Mathematics, Fibonacci n-Step

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 5-step sequences), A106309 (primes that yield a simple orbit structure in 5-step 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

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Last modified October 2 23:51 EDT 2022. Contains 357230 sequences. (Running on oeis4.)