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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.

%I A371149 #11 Mar 23 2024 05:04:41
%S A371149 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,
%T A371149 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,
%U A371149 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
%N 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.
%H A371149 Pontus von Brömssen, <a href="/A371149/b371149.txt">Table of n, a(n) for n = 2..10000</a>
%F A371149 A362333(n) = ceiling(A371148(n)/a(n)).
%e A371149 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.
%o A371149 (Python)
%o A371149 from sympy import factorint, Rational
%o A371149 def A371149(n):
%o A371149     f = factorint(n)
%o A371149     gpf = max(f, default=None)
%o A371149     a = 0
%o A371149     for p in f:
%o A371149         m = gpf
%o A371149         v = 0
%o A371149         while m >= p:
%o A371149             m //= p
%o A371149             v += m
%o A371149         a = max(a, Rational(f[p], v))
%o A371149     return a.q
%Y A371149 Cf. A000040, A006530, A362333, A371148 (numerators), A371150, A371151.
%K A371149 nonn,frac
%O A371149 2,79
%A A371149 _Pontus von Brömssen_, Mar 13 2024