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A106302
Period of the Fibonacci 3-step sequence A000073 mod prime(n).
7
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
1,1
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
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
FORMULA
a(n) = A046738(prime(n)).
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}]
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, May 02 2005, Sep 18 2008
STATUS
approved