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

1,2

Comments

This sequence differs from the corresponding Lucas sequence (A106297) 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

Chai Wah Wu, Table of n, a(n) for n = 1..388

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. 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

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}]

Prog

(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

Crossrefs

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

Sequence in context: A302911 A174481 A306205 * A157518 A001526 A295940

Adjacent sequences: A106300 A106301 A106302 * A106304 A106305 A106306

Keyword

nonn

Author

T. D. Noe, May 02 2005

Status

approved