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%I #39 Sep 16 2023 05:32:16
%S 1,2,3,2,5,6,7,1,3,10,11,6,13,14,15,2,17,6,19,10,21,22,23,3,5,26,1,14,
%T 29,30,31,2,33,34,35,6,37,38,39,5,41,42,43,22,15,46,47,6,7,10,51,26,
%U 53,2,55,7,57,58,59,30,61,62,21,1,65,66,67,34,69,70,71,3,73,74,15,38,77
%N Powerfree kernel of cubefree part of n.
%H Amiram Eldar, <a href="/A056552/b056552.txt">Table of n, a(n) for n = 1..10000</a>
%H Henry Bottomley, <a href="http://fs.gallup.unm.edu/Bottomley-Sm-Mult-Functions.htm">Some Smarandache-type multiplicative sequences</a>.
%F a(n) = A007947(A050985(n)) = A019555(A050985(n)) = n/(A053150(n)*A000189(n)) = A019555(n)/A053150(n) = A056551(n)^(1/3).
%F If n = Product_{j} Pj^Ej then a(n) = Product_{j} Pj^Fj, where Fj = 0 if Ej is 0 or a multiple of 3 and Fj = 1 otherwise.
%F Multiplicative with a(p^e) = p^(if 3|e, then 0, else 1). - _Mitch Harris_, Apr 19 2005
%F Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(6)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.3480772773... . - _Amiram Eldar_, Oct 28 2022
%F Dirichlet g.f.: zeta(3*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-1)). - _Amiram Eldar_, Sep 16 2023
%e a(32) = 2 because cubefree part of 32 is 4 and powerfree kernel of 4 is 2.
%t f[p_, e_] := p^If[Divisible[e, 3], 0, 1]; a[n_] := Times @@ (f @@@ FactorInteger[ n]); Array[a, 100] (* _Amiram Eldar_, Aug 29 2019 *)
%o (PARI) a(n) = my(f=factor(n)); for (k=1, #f~, if (frac(f[k,2]/3), f[k,2] = 1, f[k,2] = 0)); factorback(f); \\ _Michel Marcus_, Feb 28 2019
%Y Cf. A000189, A000578, A007947, A008834, A013664, A019555, A048798, A050985, A053149, A053150, A056551.
%K nonn,easy,mult
%O 1,2
%A _Henry Bottomley_, Jun 25 2000
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