A365790 a(n) = number of k <= b(n) such that rad(k) | b(n), where rad(n) = A007947(n) and b(n) = A126706(n).

8, 10, 8, 11, 8, 14, 11, 9, 8, 15, 12, 9, 16, 11, 26, 8, 10, 18, 9, 10, 14, 28, 11, 32, 10, 20, 13, 8, 15, 11, 21, 14, 10, 8, 36, 10, 33, 31, 12, 12, 27, 23, 10, 11, 41, 12, 8, 31, 18, 24, 11, 38, 8, 11, 8, 14, 44, 12, 11, 11, 25, 16, 36, 19, 33, 8, 14, 11, 26

Offset: 1

Author

Michael De Vlieger, Sep 21 2023

Keywords

• nonn

Comments

Alternatively, position of A126706(n) in the list R(rad(n)) of k such that rad(k) | n, where rad(n) = A007947(n). Note that rad(b(n)) < b(n) for all n.

Let prime p divide n. The set R(rad(n)) is a list of numbers beginning with the empty product 1 and including all k such that p | k implies p | rad(n). For example, R(6) = A003586. All k in A003586 are such that no prime q coprime to 6 divides k.

Examples

			a(1) = 8 since rad(b(1)) = rad(12) = 6, and in the sequence R(6) = A003586 = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, ...}, 12 is the 8th term.
a(2) = 10 since rad(b(2)) = rad(18) = 6, and 18 is the 10th term in R(6).
a(3) = 8 since rad(b(3)) = rad(20) = 10, and in the sequence R(10) = A003592 = {1, 2, 4, 5, 8, 10, 16, 20, ...}, 20 is the 8th term.
a(4) = 11 since rad(b(4)) = rad(24) = 6, and 24 is the 11th term in R(6).
a(5) = 8 since rad(b(5)) = rad(28) = 14, and in the sequence R(14) = A003591 = {1, 2, 4, 7, 8, 14, 16, 28, ...}, 28 is the 8th term, etc.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007947, A010846, A126706, A365783, A365791.

Programs

Mathematica

nn = 220;
f[x_] := f[x] = Times @@ FactorInteger[x][[All, 1]];
t = Select[Range[nn], Nor[PrimePowerQ[#], SquareFreeQ[#]] &];
s = Map[f, t];
Map[Function[k, Set[r[k], Select[Range[nn], Divisible[k, f[#]] &]]], Union@ s];
Array[FirstPosition[r[s[[#]]], t[[#]] ][[1]] &, Length[t] ]

Formula

a(n) = A010846(A126706(n)).