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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.
5

%I #12 Mar 23 2024 05:04:26

%S 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 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 1,1,3,1,1,2,1,1,1,1,4,4,1,1,1,1,1,1,1

%N 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 Pontus von Brömssen, <a href="/A371148/b371148.txt">Table of n, a(n) for n = 2..10000</a>

%F A362333(n) = ceiling(a(n)/A371149(n)).

%e 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 (Python)

%o from sympy import factorint,Rational

%o def A371148(n):

%o f = factorint(n)

%o gpf = max(f,default=None)

%o a = 0

%o for p in f:

%o m = gpf

%o v = 0

%o while m >= p:

%o m //= p

%o v += m

%o a = max(a,Rational(f[p],v))

%o return a.p

%Y Cf. A000040, A006530, A362333, A371149 (denominators), A371150, A371151.

%K nonn,frac

%O 2,3

%A _Pontus von Brömssen_, Mar 13 2024