OFFSET
1,8
COMMENTS
a(n) = 0 for all odd values n > 3 since the Pisano period of m is always even except when m=1 or 2.
The Pisano period of m divides n if and only if F_m = 0 (mod n) and F_{m+1} = 1 (mod n), hence n | gcd(F_m,F_{m+1}-1).
Conjecture: If n is of the form 12*5^j (i.e. 2n is of the form 24*5^j which has the unique property that pi(2n)=2n), then a(n) > a(m) for all m < n.
Conjecture: Every natural number appears on this list.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..314 (terms 1..239 from Oliver Lippard)
B. Benfield and O. Lippard, Fixed points of K-Fibonacci Pisano periods, arXiv:2404.08194 [math.NT], 2024.
J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithmetica 16 (1969), 105-110.
EXAMPLE
a(9) = 3 because the Pisano periods of 76, 38, and 19, but no others, are 2*9=18.
There are no numbers with Pisano period 2 or 4, so a(1) = a(2) = 0.
PROG
(Sage)
def a(n):
num=0
if n<3:
return 0
x=gcd(fibonacci(2*n), fibonacci(2*n+1)-1)
for d in divisors(x):
if BinaryRecurrenceSequence(1, 1, 0, 1).period(d)==2*n:
num+=1
return num
for i in range(1, 101):
print(a(i))
(Python)
from functools import lru_cache
from math import gcd, lcm
from sympy import factorint, divisors, fibonacci
def A375089(n):
@lru_cache(maxsize=None)
def A001175(n):
if n == 1:
return 1
f = factorint(n).items()
if len(f) > 1:
return lcm(*(A001175(a**b) for a, b in f))
else:
k, x = 1, (1, 1)
while x != (0, 1):
k += 1
x = (x[1], (x[0]+x[1]) % n)
return k
a, b = fibonacci((m:=n<<1)+1), fibonacci(m)
return sum(1 for d in divisors(gcd(a-1, b), generator=True) if A001175(d)==m) # Chai Wah Wu, Aug 28 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Oliver Lippard and Brennan G. Benfield, Jul 29 2024
STATUS
approved