A353693 - OEIS

%I #17 May 07 2022 06:49:57

%S 1,1,1,1,1,2,1,1,1,2,1,1,1,2,3,1,1,1,1,1,3,2,1,1,1,2,1,1,1,12,1,1,3,2,

%T 5,2,1,2,3,1,1,12,1,1,1,2,1,1,1,1,3,1,1,1,5,1,3,2,1,6,1,2,1,1,5,12,1,

%U 1,3,20,1,1,1,2,1,1,7,12,1,1,1,2,1,6,5,2

%N a(n) is the least multiplier k such that the exponents in the prime factorization of k*n are mutually distinct (A130091).

%C First differs from A327499 at n = 30.

%C If n = Product_{i=1..k} p_i is squarefree (A005117), and p_1 < p_2 < ... < p_k are its k ordered prime divisors, then a(n) = Product_{i} p_i^(k-i).

%C If n is powerful (A001694) then a(n) = a(n/rad(n)), where rad(n) is the squarefree kernel of n (A007947). In general, if k = A051904(n) is the minimal exponent in the prime factorization of n, then a(n) = a(n/(rad(n)^(k-1))).

%H Amiram Eldar, <a href="/A353693/b353693.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = 1 if and only if n is in A130091.

%F a(A130092(n)) > 1.

%F rad(a(n)) | rad(n).

%F a(n) = A353694(n)/n.

%e a(2) = 1 since 2 = 2^1 has only one exponent (1) in its prime factorization.

%e a(6) = 2 since 6 = 2*3 has two equal exponents (1) in its prime factorization, and 2*6 = 12 = 2^2*3 has two distinct exponents (1 and 2).

%t a[n_] := Module[{k = 1}, While[!UnsameQ @@ FactorInteger[k*n][[;; , 2]], k++]; k]; Array[a, 100]

%o (PARI) a(n) = my(k=1, f=factor(n)[,2]); while(#Set(f) != #f, k++; f=factor(k*n)[,2]); k; \\ _Michel Marcus_, May 05 2022

%Y Cf. A001694, A005117, A007947, A130091, A130092, A327498, A327499, A353694.

%K nonn

%O 1,6

%A _Amiram Eldar_, May 04 2022