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If n = Product (p_j^k_j) then a(n) = Product (prime(p_j)).
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%I #6 Oct 29 2019 12:15:25

%S 1,3,5,3,11,15,17,3,5,33,31,15,41,51,55,3,59,15,67,33,85,93,83,15,11,

%T 123,5,51,109,165,127,3,155,177,187,15,157,201,205,33,179,255,191,93,

%U 55,249,211,15,17,33,295,123,241,15,341,51,335,327,277,165,283,381,85,3,451

%N If n = Product (p_j^k_j) then a(n) = Product (prime(p_j)).

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>

%e a(54) = 15 because 54 = 2 * 3^3 = prime(1) * prime(2)^3 and prime(prime(1)) * prime(prime(2)) = 3 * 5 = 15.

%t a[n_] := Times @@ (Prime[#[[1]]] & /@ FactorInteger[n]); Table[a[n], {n, 1, 65}]

%o (PARI) a(n)={my(f=factor(n)[,1]); prod(i=1, #f, prime(f[i]))} \\ _Andrew Howroyd_, Oct 29 2019

%Y Cf. A006450, A007947, A064988, A156061, A181819, A321874.

%K nonn,mult

%O 1,2

%A _Ilya Gutkovskiy_, Oct 29 2019