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%I #26 Oct 27 2022 07:32:00
%S 1,2,3,2,5,6,7,2,3,10,11,6,13,14,15,1,17,6,19,10,21,22,23,6,5,26,3,14,
%T 29,30,31,2,33,34,35,6,37,38,39,10,41,42,43,22,15,46,47,3,7,10,51,26,
%U 53,6,55,14,57,58,59,30,61,62,21,2,65,66,67,34,69,70,71,6,73,74,15,38
%N Powerfree kernel of 4th-powerfree part of n.
%H Amiram Eldar, <a href="/A056554/b056554.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(A053165(n)) = A053166(A053165(n)) = n/(A053164(n)*A000190(n)) = A053166(n)/A053164(n) = A056553(n)^(1/4).
%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 4 and Fj = 1 otherwise.
%F Multiplicative with a(p^e) = p^(if 4|e, then 0, else 1). - _Mitch Harris_, Apr 19 2005
%F Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(8)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.3513111135... . - _Amiram Eldar_, Oct 27 2022
%e a(64) = 2 because 4th-power-free part of 64 is 4 and power-free kernel of 4 is 2.
%t f[p_, e_] := p^If[Divisible[e, 4], 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]/4), f[k,2] = 1, f[k,2] = 0)); factorback(f); \\ _Michel Marcus_, Feb 28 2019
%Y Cf. A000190, A007947, A008835, A013666, A053164, A053165, A053166, A053167, A056552, A056553, A056555.
%K nonn,mult
%O 1,2
%A _Henry Bottomley_, Jun 25 2000
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