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A362333
Least nonnegative integer k such that (gpf(n)!)^k is divisible by n, where gpf(n) is the greatest prime factor of n.
5
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
1,4
COMMENTS
First differs from A088388 at n = 40.
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
EXAMPLE
For n = 12, gpf(n)! = 3! = 6 is not divisible by 12, but (3!)^2 = 36 is divisible by 12, so a(12) = 2.
PROG
(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
KEYWORD
nonn
AUTHOR
STATUS
approved