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A371148 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 numerator of the maximum of e_j/f_j.

%I A371148 #12 Mar 23 2024 05:04:26
%S A371148 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,
%T A371148 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,
%U A371148 1,1,3,1,1,2,1,1,1,1,4,4,1,1,1,1,1,1,1
%N A371148 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 numerator of the maximum of e_j/f_j.
%H A371148 Pontus von Brömssen, <a href="/A371148/b371148.txt">Table of n, a(n) for n = 2..10000</a>
%F A371148 A362333(n) = ceiling(a(n)/A371149(n)).
%e A371148 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) = 4.
%o A371148 (Python)
%o A371148 from sympy import factorint,Rational
%o A371148 def A371148(n):
%o A371148     f = factorint(n)
%o A371148     gpf = max(f,default=None)
%o A371148     a = 0
%o A371148     for p in f:
%o A371148         m = gpf
%o A371148         v = 0
%o A371148         while m >= p:
%o A371148             m //= p
%o A371148             v += m
%o A371148         a = max(a,Rational(f[p],v))
%o A371148     return a.p
%Y A371148 Cf. A000040, A006530, A362333, A371149 (denominators), A371150, A371151.
%K A371148 nonn,frac
%O A371148 2,3
%A A371148 _Pontus von Brömssen_, Mar 13 2024