A106297 Period of the Lucas 5-step sequence A074048 mod n.

1, 1, 104, 6, 781, 104, 2801, 12, 312, 781, 16105, 312, 30941, 2801, 81224, 24, 88741, 312, 13032, 4686, 291304, 16105, 12166, 312, 3905, 30941, 936, 16806, 70728, 81224, 190861, 48, 1674920, 88741, 2187581, 312, 1926221, 13032, 3217864, 9372, 2896405

Offset

1,3

Comments

This sequence differs from the corresponding Fibonacci sequence (A106303) at all n that are multiples of 2 or 599 because 9584 is the discriminant of the characteristic polynomial x^5-x^4-x^3-x^2-x-1 and the prime factors of 9584 are 2 and 599.

Links

Table of n, a(n) for n=1..41.

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

Formula

Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).

Conjectures: a(5^k) = 781*5^(k-1) for k > 0. a(2^k) = 3*2^(k-1) for k > 1. If a(p) != a(p^2) for prime p > 2, then a(p^k) = p^(k-1)*a(p) for k > 0. - Chai Wah Wu, Feb 25 2022

Mathematica

n=5; Table[p=i; a=Join[Table[ -1, {n-1}], {n}]; a=Mod[a, p]; a0=a; k=0; While[k++; s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; a!=a0]; k, {i, 50}]

Prog

(Python)
from itertools import count
def A106297(n):
    a = b = (5%n, 1%n, 7%n, 3%n, 15%n)
    s = sum(b) % n
    for m in count(1):
        b, s = b[1:] + (s, ), (s+s-b[0]) % n
        if a == b:
            return m # Chai Wah Wu, Feb 22-27 2022

Crossrefs

Cf. A074048, A106303 (period of Fibonacci 5-step sequence mod n), A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).

Sequence in context: A088584 A238490 A097014 * A090849 A091025 A054904

Adjacent sequences: A106294 A106295 A106296 * A106298 A106299 A106300

Keyword

nonn

Author

T. D. Noe, May 02 2005, Nov 19 2006

Status

approved