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A285769
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(Product of distinct prime factors)^(Product of prime exponents).
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5
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 36, 13, 14, 15, 16, 17, 36, 19, 100, 21, 22, 23, 216, 25, 26, 27, 196, 29, 30, 31, 32, 33, 34, 35, 1296, 37, 38, 39, 1000, 41, 42, 43, 484, 225, 46, 47, 1296, 49, 100, 51, 676, 53, 216, 55, 2744, 57, 58, 59, 900, 61, 62, 441
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OFFSET
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1,2
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COMMENTS
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a(n) and A066638 differ at {36, 72, 100, 108, 144, ...}, i.e., for all n in A036785, since a(n) takes the product of the multiplicities of prime factors of n, while A066638 takes the maximum value of the multiplicities of prime factors of n. For these n, a(n) > A066638(n).
a(1) = 1 since 1 is the empty product; 1^1 = 1.
a(p) = p since omega(p) = A001221(p) = 1 thus p^1 = p.
a(p^m) = p^m since omega(p) = 1 thus p^m is maintained.
For squarefree n with omega(n) > 1, a(n) = n.
For n with omega(n) > 1 and at least one multiplicity m > 1, a(n) > n. In other words, let a(n) = k^m, where k is the product of the distinct prime factors of n and m is the product of the multiplicities of the distinct prime factors of n. a(n) > n for n in A126706 since there are 2 or more prime factors in k and m > 1.
Squarefree kernels of terms a(n) > n: {6, 6, 10, 6, 14, 6, 10, 22, 15, 6, 10, 26, 6, 14, 30, 21, ...}.
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LINKS
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FORMULA
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EXAMPLE
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a(2) = 2 since (2)^(1) = 2^1 = 2.
a(6) = 6 since (2*3)^(1*1) = 6^1 = 6.
a(12) = 36 since (2*3)^(2*1) = 6^2 = 36.
a(30) = 30 since (2*3*5)^(1*1*1) = 30^1 = 30.
a(144) = 1679616 since (2*3)^(4*2) = 6^8 = 1679616.
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MATHEMATICA
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Array[Power @@ Map[Times @@ # &, Transpose@ FactorInteger@ #] &, 63] (* Michael De Vlieger, Apr 25 2017 *)
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PROG
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(Python)
from sympy import divisor_count, divisors
from sympy.ntheory.factor_ import core
def rad(n): return max(list(filter(lambda i: core(i) == i, divisors(n))))
def a(n): return rad(n)**divisor_count(n/rad(n)) # Indranil Ghosh, Apr 26 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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