A362333 Least nonnegative integer k such that (gpf(n)!)^k is divisible by n, where gpf(n) is the greatest prime factor of n.

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 1, 1, 1, 1, 3, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 4, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Pontus von Brömssen, Apr 16 2023

Keywords

• nonn

Comments

First differs from A088388 at n = 40.

Examples

			For n = 12, gpf(n)! = 3! = 6 is not divisible by 12, but (3!)^2 = 36 is divisible by 12, so a(12) = 2.
		

Crossrefs

Cf. A006530, A051903, A057109, A088388, A371148, A371149, A371151, A371152.

Programs

Python

from sympy import factorint
def A362333(n):
    f = factorint(n)
    gpf = max(f,default=None)
    a = 0
    for p in f:
        m = gpf
        v = 0
        while m >= p:
            m //= p
            v += m
        a = max(a,-(-f[p]//v))
    return a

Formula

a(n) > 1 if and only if n is in A057109.

a(n) <= A051903(n).

a(n) = ceiling(A371148(n)/A371149(n)). - Pontus von Brömssen, Mar 16 2024