A365790 - OEIS

%I #5 Oct 06 2023 11:10:03

%S 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,

%T 20,13,8,15,11,21,14,10,8,36,10,33,31,12,12,27,23,10,11,41,12,8,31,18,

%U 24,11,38,8,11,8,14,44,12,11,11,25,16,36,19,33,8,14,11,26

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

%C 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.

%C 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.

%H Michael De Vlieger, <a href="/A365790/b365790.txt">Table of n, a(n) for n = 1..10000</a>

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

%e 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.

%e a(2) = 10 since rad(b(2)) = rad(18) = 6, and 18 is the 10th term in R(6).

%e 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.

%e a(4) = 11 since rad(b(4)) = rad(24) = 6, and 24 is the 11th term in R(6).

%e 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.

%t nn = 220;

%t f[x_] := f[x] = Times @@ FactorInteger[x][[All, 1]];

%t t = Select[Range[nn], Nor[PrimePowerQ[#], SquareFreeQ[#]] &];

%t s = Map[f, t];

%t Map[Function[k, Set[r[k], Select[Range[nn], Divisible[k, f[#]] &]]], Union@ s];

%t Array[FirstPosition[r[s[[#]]], t[[#]] ][[1]] &, Length[t] ]

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

%K nonn

%O 1,1

%A _Michael De Vlieger_, Sep 21 2023