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A162947
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Numbers k such that the product of all divisors of k equals k^3.
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4
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1, 12, 18, 20, 28, 32, 44, 45, 50, 52, 63, 68, 75, 76, 92, 98, 99, 116, 117, 124, 147, 148, 153, 164, 171, 172, 175, 188, 207, 212, 236, 242, 243, 244, 245, 261, 268, 275, 279, 284, 292, 316, 325, 332, 333, 338, 356, 363, 369, 387, 388, 404, 412, 423, 425, 428
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OFFSET
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1,2
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COMMENTS
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Contains the terms of A054753 (products p*q^2 of a prime p and the square of a different prime q), 1, and p^5, where p is prime.
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LINKS
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FORMULA
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EXAMPLE
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18 is in the sequence because the product of its divisors is 1 * 2 * 3 * 6 * 9 * 18 = 18^3.
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MATHEMATICA
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Select[Range[500], Surd[Times@@Divisors[#], 3] == # &] (* Harvey P. Dale, Mar 15 2017 *)
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PROG
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(PARI) isok(n) = my(d = divisors(n)); prod(i=1, #d, d[i]) == n^3; \\ Michel Marcus, Feb 04 2014
(Python)
from itertools import chain, count, islice
from sympy import divisor_count
def A162947_gen(): # generator of terms
return chain((1, ), filter(lambda n:divisor_count(n)==6, count(2)))
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CROSSREFS
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Cf. A008578 (product of divisors equals n), A007422 (product of divisors equals n^2).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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