A106302 Period of the Fibonacci 3-step sequence A000073 mod prime(n).

4, 13, 31, 48, 110, 168, 96, 360, 553, 140, 331, 469, 560, 308, 46, 52, 3541, 1860, 1519, 5113, 5328, 3120, 287, 8011, 3169, 680, 51, 1272, 990, 12883, 5376, 5720, 18907, 3864, 7400, 2850, 8269, 162, 9296, 2494, 32221, 10981, 36673, 4656, 3234, 198, 5565

Offset: 1

Author

T. D. Noe, May 02 2005, Sep 18 2008

Keywords

• nonn

Comments

This sequence differs from the corresponding Lucas sequence (A106294) at n=1 and 5 because these correspond to the primes 2 and 11, which are the prime factors of -44, the discriminant of the characteristic polynomial x^3-x^2-x-1. We have a(n) < prime(n) for the primes in A106279.

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, Fibonacci n-Step Number

Crossrefs

Cf. A106273, A106279, A106294, A046738.

Programs

Mathematica

n=3; Table[p=Prime[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,60}]

Python

from itertools import count
from sympy import prime
def A106302(n):
    a = b = (0,)*2+(1 % (p:= prime(n)),)
    for m in count(1):
        b = b[1:] + (sum(b) % p,)
        if a == b:
            return m # Chai Wah Wu, Feb 27 2022

Formula

a(n) = A046738(prime(n)).