A375519 Number of positive integers with Pisano period equal to n.

1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 3, 0, 2, 0, 1, 0, 10, 0, 1, 0, 2, 0, 8, 0, 4, 0, 1, 0, 9, 0, 1, 0, 11, 0, 8, 0, 2, 0, 3, 0, 55, 0, 6, 0, 2, 0, 6, 0, 11, 0, 3, 0, 49, 0, 1, 0, 8, 0, 8, 0, 2, 0, 13, 0, 133, 0, 1, 0, 6, 0, 20, 0, 46, 0, 1, 0, 49, 0, 3, 0, 27, 0, 81, 0, 4, 0, 10, 260, 0, 2, 0, 38

Offset: 1

Author

Oliver Lippard and Brennan G. Benfield, Jul 29 2024

Keywords

• nonn

Comments

See also A375089, which is the main entry.

[This was the version of A375089 that was originally submitted. I think both belong in the OEIS. - N. J. A. Sloane, Aug 28 2024]

Crossrefs

Cf. A001175, A375089.

Programs

Python

from functools import lru_cache
from math import gcd, lcm
from sympy import factorint, divisors, fibonacci
def A375519(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(n+1), fibonacci(n)
    return sum(1 for d in divisors(gcd(a-1,b),generator=True) if A001175(d)==n) # Chai Wah Wu, Aug 28 2024

Formula

a(2n) = A375089(n). - Chai Wah Wu, Aug 28 2024