A371149 Let n = Product_{j=1..k} p_j^e_j and gpf(n)! = Product_{j=1..k} p_j^f_j, where p_j = A000040(j) is the j-th prime and p_k = gpf(n) = A006530(n) is the greatest prime factor of n. a(n) is the denominator of the maximum of e_j/f_j.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 2

Author

Pontus von Brömssen, Mar 13 2024

Keywords

• nonn
• frac

Examples

			For n = 80 = 2^4 * 3^0 * 5^1, gpf(80)! = 5! = 2^3 * 3^1 * 5^1. The ratios of the prime exponents are 4/3, 0/1, and 1/1, the greatest of which is 4/3, so a(80) = 3.
		

Links

• Pontus von Brömssen, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A000040, A006530, A362333, A371148 (numerators), A371150, A371151.

Programs

Python

from sympy import factorint, Rational
def A371149(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, Rational(f[p], v))
    return a.q

Formula

A362333(n) = ceiling(A371148(n)/a(n)).