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A362972 Squarefree kernels of cubefull numbers (A036966).

%I A362972 #22 Mar 27 2025 05:43:33
%S A362972 1,2,2,3,2,2,3,5,2,6,3,2,7,6,2,5,6,3,6,10,2,6,11,6,6,10,2,3,13,7,6,14,
%T A362972 5,15,6,6,10,2,17,10,6,14,6,3,19,6,6,10,2,21,10,15,6,22,14,6,23,6,11,
%U A362972 6,5,10,2,7,15,6,26,14,3,10,6,22,14,6,29,10,30,6
%N A362972 Squarefree kernels of cubefull numbers (A036966).
%H A362972 Amiram Eldar, <a href="/A362972/b362972.txt">Table of n, a(n) for n = 1..10000</a>
%H A362972 Rafael Jakimczuk, <a href="https://doi.org/10.12988/imf.2017.7759">The kernel of powerful numbers</a>, International Mathematical Forum, Vol. 12, No. 15 (2017), pp. 721-730, eq. (18), p. 728.
%F A362972 a(n) = A007947(A036966(n)).
%F A362972 Sum_{A036966(k) < x} a(k) = c * x^(2/3) + o(x^(2/3)), where c = (3/Pi^2) * Product_{p prime} (1 + 1/((p+1)*(p^(2/3)-1))) = 0.7356919531... (Jakimczuk, 2017). [corrected Sep 21 2024]
%F A362972 Sum_{k=1..n} a(k) ~ (c / A362974 ^ 2) * n^2, where c is the constant above.
%e A362972 A036966(2) = 8 = 2^3, therefore a(2) = 2.
%e A362972 A036966(10) = 216 = 2^3 * 3^2, therefore a(10) = 2 * 3 = 6.
%t A362972 seq[kmax_] := Module[{s = {1}}, Do[f = FactorInteger[k]; If[Min@f[[;; , 2]] > 2, AppendTo[s, Times @@ f[[;; , 1]]]], {k, 2, kmax}]; s]; seq[10^5]
%o A362972 (PARI) lista(kmax) = {my(f); for(k = 1, kmax, f = factor(k); if(k==1 || vecmin(f[, 2]) > 2, print1(vecprod(f[, 1]), ", ")));}
%Y A362972 Cf. A007947, A036966, A084371, A362974.
%K A362972 nonn
%O A362972 1,2
%A A362972 _Amiram Eldar_, May 13 2023