A375519 -id:A375519 - cp's OEIS frontend

0, 0, 1, 1, 1, 1, 1, 2, 3, 2, 1, 10, 1, 2, 8, 4, 1, 9, 1, 11, 8, 2, 3, 55, 6, 2, 6, 11, 3, 49, 1, 8, 8, 2, 13, 133, 1, 6, 20, 46, 1, 49, 3, 27, 81, 4, 1, 260, 2, 38, 20, 11, 1, 106, 21, 78, 20, 4, 7, 874, 1, 6, 81, 48, 29, 49, 3, 27, 42, 108, 1, 1319, 3, 14, 174, 23, 13, 101, 1, 444

Offset: 1

Oliver Lippard and Brennan G. Benfield, Jul 29 2024

nonn

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.

			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.

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.

Cf. A000045, A001175.

See A375519 for the "equal to n" version.

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

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))

cp's OEIS Frontend

A375089 Number of positive integers with Pisano period equal to 2n.

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