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%I #26 Jun 25 2022 21:54:02
%S 1,12,18,20,28,32,44,45,50,52,63,68,75,76,92,98,99,116,117,124,147,
%T 148,153,164,171,172,175,188,207,212,236,242,243,244,245,261,268,275,
%U 279,284,292,316,325,332,333,338,356,363,369,387,388,404,412,423,425,428
%N Numbers k such that the product of all divisors of k equals k^3.
%C 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.
%H Harvey P. Dale, <a href="/A162947/b162947.txt">Table of n, a(n) for n = 1..1000</a>
%F {n: A007955(n) = A000578(n)}. - _R. J. Mathar_, Jul 19 2009
%F {1} UNION A030515. - _R. J. Mathar_, Jul 19 2009
%F Numbers k such that k^2 = product of proper divisors of k. - _Juri-Stepan Gerasimov_, May 03 2011
%e 18 is in the sequence because the product of its divisors is 1 * 2 * 3 * 6 * 9 * 18 = 18^3.
%t Select[Range[500], Surd[Times@@Divisors[#], 3] == # &] (* _Harvey P. Dale_, Mar 15 2017 *)
%o (PARI) isok(n) = my(d = divisors(n)); prod(i=1, #d, d[i]) == n^3; \\ _Michel Marcus_, Feb 04 2014
%o (Python)
%o from itertools import chain, count, islice
%o from sympy import divisor_count
%o def A162947_gen(): # generator of terms
%o return chain((1,),filter(lambda n:divisor_count(n)==6,count(2)))
%o A162947_list = list(islice(A162947_gen(),20)) # _Chai Wah Wu_, Jun 25 2022
%Y Cf. A111398, A030628. - _R. J. Mathar_, Jul 19 2009
%Y Cf. A008578 (product of divisors equals n), A007422 (product of divisors equals n^2).
%K nonn
%O 1,2
%A _Claudio Meller_, Jul 18 2009
%E Edited by _R. J. Mathar_, Jul 19 2009
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