OFFSET
1,1
COMMENTS
Also called cubeful numbers, but this term is ambiguous and is best avoided.
Numbers n such that A007427(n) = sum(d|n,mu(d)*mu(n/d)) == 0. - Benoit Cloitre, Apr 17 2002
The convention in the OEIS is that squareful, cubeful, biquadrateful (A046101), ... mean the same as "not squarefree" etc., while 2- or square-full, 3- or cube-full (A036966), 4-full (A036967) are used for Golomb's notion of powerful numbers (A001694, see references there), when each prime factor occurs to a power > 1. - M. F. Hasler, Feb 12 2008. Added by N. J. A. Sloane, Apr 25 2023: This suggestion has not been a success. It is hopeless to try to make a distinction between "cubeful" and "cubefull". To avoid ambiguity, do not use either term, but instead say exactly what you mean.
Also solutions to equation tau_{-2}(n)=0, where tau_{-2} is A007427. - Enrique PĂ©rez Herrero, Jan 19 2013
The asymptotic density of this sequence is 1 - 1/zeta(3) = 0.168092... - Amiram Eldar, Jul 09 2020
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Cubefree.
FORMULA
A212793(a(n)) = 0. - Reinhard Zumkeller, May 27 2012
Sum_{n>=1} 1/a(n)^s = (zeta(s)*(zeta(3*s)-1))/zeta(3*s). - Amiram Eldar, Dec 27 2022
MAPLE
isA046099 := proc(n)
local p;
for p in ifactors(n)[2] do
if op(2, p) >= 3 then
return true;
end if;
end do:
false ;
end proc:
for n from 1 do
if isA046099(n) then
printf("%d\n", n) ;
end if;
end do: # R. J. Mathar, Dec 08 2015
MATHEMATICA
lst={}; Do[a=0; Do[If[FactorInteger[m][[n, 2]]>2, a=1], {n, Length[FactorInteger[m]]}]; If[a==1, AppendTo[lst, m]], {m, 10^3}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 15 2008 *)
PROG
(Haskell)
a046099 n = a046099_list !! (n-1)
a046099_list = filter ((== 1) . a212793) [1..]
-- Reinhard Zumkeller, May 27 2012
(PARI) is(n)=n>7 && vecmax(factor(n)[, 2])>2 \\ Charles R Greathouse IV, Sep 17 2015
(Python)
from sympy.ntheory.factor_ import core
def ok(n): return core(n, 3) != n
print(list(filter(ok, range(1, 337)))) # Michael S. Branicky, Aug 16 2021
(Python)
from sympy import mobius, integer_nthroot
def A046099(n):
def f(x): return n+sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x, 3)[0]+1))
m, k = n, f(n)
while m != k:
m, k = k, f(k)
return m # Chai Wah Wu, Aug 05 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
More terms from Vladimir Joseph Stephan Orlovsky, Aug 15 2008
Edited by N. J. A. Sloane, Jul 27 2009
STATUS
approved