A106298 Period of the Lucas 5-step sequence A074048 mod prime(n).

1, 104, 781, 2801, 16105, 30941, 88741, 13032, 12166, 70728, 190861, 1926221, 2896405, 79506, 736, 8042221, 102689, 3720, 20151120, 2863280, 546120, 39449441, 48030024, 3690720, 29509760, 104060400, 37516960, 132316201, 28231632, 6384, 86714880, 2248090, 3128

Offset: 1

Author

T. D. Noe, May 02 2005

Keywords

• nonn

Comments

This sequence is the same as the period of Fibonacci 5-step sequence (A106304) mod prime(n) except for n=1 and 109, which correspond to the primes 2 and 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. We have a(n) < prime(n) for the primes 2, 599 and A106281.

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..82
• Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Crossrefs

Cf. A106281 (primes p such that x^5-x^4-x^3-x^2-x-1 mod p has 5 distinct zeros), A106297.

Programs

Mathematica

n=5; Table[p=Prime[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, 40}]

Python

from itertools import count
from sympy import prime
def A106298(n):
    a = b = (5%(p:=prime(n)),1%p,7%p,3%p,15%p)
    s = sum(b) % p
    for m in count(1):
        b, s = b[1:] + (s,), (s+s-b[0]) % p
        if a == b:
            return m # Chai Wah Wu, Feb 22-27 2022

Formula

a(n) = A106297(prime(n)).

Extensions

a(31)-a(33) from Chai Wah Wu, Feb 27 2022