A162947 Numbers k such that the product of all divisors of k equals k^3.

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

Offset

1,2

Comments

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.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Formula

{n: A007955(n) = A000578(n)}. - R. J. Mathar, Jul 19 2009

{1} UNION A030515. - R. J. Mathar, Jul 19 2009

Numbers k such that k^2 = product of proper divisors of k. - Juri-Stepan Gerasimov, May 03 2011

Example

18 is in the sequence because the product of its divisors is 1 * 2 * 3 * 6 * 9 * 18 = 18^3.

Mathematica

Select[Range[500], Surd[Times@@Divisors[#], 3] == # &] (* Harvey P. Dale, Mar 15 2017 *)

Prog

(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)))
A162947_list = list(islice(A162947_gen(), 20)) # Chai Wah Wu, Jun 25 2022

Crossrefs

Cf. A111398, A030628. - R. J. Mathar, Jul 19 2009

Cf. A008578 (product of divisors equals n), A007422 (product of divisors equals n^2).

Sequence in context: A217856 A253388 A030515 * A351201 A359892 A070011

Adjacent sequences: A162944 A162945 A162946 * A162948 A162949 A162950

Keyword

nonn

Author

Claudio Meller, Jul 18 2009

Extensions

Edited by R. J. Mathar, Jul 19 2009

Status

approved