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A371148
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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.
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5
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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, 4, 4, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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2,3
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LINKS
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FORMULA
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EXAMPLE
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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.
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PROG
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(Python)
from sympy import factorint, Rational
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.p
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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