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%I #19 Sep 16 2023 05:33:30
%S 1,8,27,8,125,216,343,1,27,1000,1331,216,2197,2744,3375,8,4913,216,
%T 6859,1000,9261,10648,12167,27,125,17576,1,2744,24389,27000,29791,8,
%U 35937,39304,42875,216,50653,54872,59319,125,68921,74088,79507,10648
%N Smallest cube divisible by n divided by largest cube which divides n.
%H Amiram Eldar, <a href="/A056551/b056551.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) = A053149(n)/A008834(n) = A048798(n)*A050985(n) = A056552(n)^3.
%F From _Amiram Eldar_, Oct 28 2022: (Start)
%F Multiplicative with a(p^e) = 1 if e is divisible by 3, and a(p^e) = p^3 otherwise.
%F Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(12)/(4*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A013670 * A330596 / (4*A002117) = 0.1557163105... . (End)
%F Dirichlet g.f.: zeta(3*s) * Product_{p prime} (1 + 1/p^(s-3) + 1/p^(2*s-3)). - _Amiram Eldar_, Sep 16 2023
%e a(16) = 8 since smallest cube divisible by 16 is 64 and smallest cube which divides 16 is 8 and 64/8 = 8.
%t f[p_, e_] := p^If[Divisible[e, 3], 0, 1]; a[n_] := (Times @@ (f @@@ FactorInteger[ n]))^3; Array[a, 100] (* _Amiram Eldar_, Aug 29 2019*)
%o (PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%3, f[i,1], 1))^3; } \\ _Amiram Eldar_, Oct 28 2022
%Y Cf. A000189, A000578, A008834, A019555, A048798, A050985, A053149, A053150, A056552.
%Y Cf. A002117, A013670, A330596.
%K nonn,easy,mult
%O 1,2
%A _Henry Bottomley_, Jun 25 2000
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