A106303 Period of the Fibonacci 5-step sequence A001591 mod n.

1, 6, 104, 12, 781, 312, 2801, 24, 312, 4686, 16105, 312, 30941, 16806, 81224, 48, 88741, 312, 13032, 9372, 291304, 96630, 12166, 312, 3905, 185646, 936, 33612, 70728, 243672, 190861, 96, 1674920, 532446, 2187581, 312, 1926221, 13032

Offset: 1

Author

T. D. Noe, May 02 2005

Keywords

• nonn
• changed

Comments

This sequence can differ from the corresponding Lucas sequence (A106297) only when n is a multiple 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. [Corrected by Avery Diep, Aug 25 2025]

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..388
• Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Crossrefs

Cf. A001591, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1), A106297 (period of Lucas 5-step sequence mod n).

Programs

Mathematica

n=5; Table[p=i; a=Join[{1}, Table[0, {n-1}]] 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}]

Python

from itertools import count
def A106303(n):
    a = b = (0,)*4+(1 % n,)
    s = 1 % 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 21-27 2022

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