A362972 Squarefree kernels of cubefull numbers (A036966).

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, 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, 6, 5, 10, 2, 7, 15, 6, 26, 14, 3, 10, 6, 22, 14, 6, 29, 10, 30, 6

Offset: 1

Author

Amiram Eldar, May 13 2023

Keywords

• nonn

Examples

			A036966(2) = 8 = 2^3, therefore a(2) = 2.
A036966(10) = 216 = 2^3 * 3^2, therefore a(10) = 2 * 3 = 6.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Rafael Jakimczuk, The kernel of powerful numbers, International Mathematical Forum, Vol. 12, No. 15 (2017), pp. 721-730, eq. (18), p. 728.

Crossrefs

Cf. A007947, A036966, A084371, A362974.

Programs

Mathematica

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]

PARI

lista(kmax) = {my(f); for(k = 1, kmax, f = factor(k); if(k==1 || vecmin(f[, 2]) > 2, print1(vecprod(f[, 1]), ", ")));}

Formula

a(n) = A007947(A036966(n)).

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]

Sum_{k=1..n} a(k) ~ (c / A362974 ^ 2) * n^2, where c is the constant above.