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A328248 a(n) = 1 if n is a squarefree number (A005117), otherwise a(n) = 1 + number of iterations of arithmetic derivative (A003415) needed to reach a squarefree number, or 0 if no such number is ever reached.

%I A328248 #14 Oct 11 2019 16:56:18
%S A328248 1,1,1,0,1,1,1,0,2,1,1,0,1,1,1,0,1,2,1,0,1,1,1,0,2,1,0,0,1,1,1,0,1,1,
%T A328248 1,0,1,1,1,0,1,1,1,0,2,1,1,0,2,3,1,0,1,0,1,0,1,1,1,0,1,1,2,0,1,1,1,0,
%U A328248 1,1,1,0,1,1,2,0,1,1,1,0,0,1,1,0,1,1,1,0,1,2,1,0,1,1,1,0,1,2,3,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,2,1,1,0
%N A328248 a(n) = 1 if n is a squarefree number (A005117), otherwise a(n) = 1 + number of iterations of arithmetic derivative (A003415) needed to reach a squarefree number, or 0 if no such number is ever reached.
%H A328248 Antti Karttunen, <a href="/A328248/b328248.txt">Table of n, a(n) for n = 1..65537</a>
%F A328248 a(4*n) = a(27*n) = 0 and in general, a(m * p^p) = 0, for any m >= 1 and any prime p.
%e A328248 For n = 9, it itself is not a squarefree number, while its arithmetic derivative A003415(9) = 6 is, so it took just one iteration to find a squarefree number, thus a(9) = 1+1 = 2.
%e A328248 For n = 50, which is not squarefree, and its first derivative A003415(50) = 45 also is not squarefree, but taking derivative yet again, gives A003415(45) = 39 = 3*13, which is squarefree, thus a(50) = 2+1 = 3.
%o A328248 (PARI)
%o A328248 A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i,2]>=f[i,1],return(0), s += f[i, 2]/f[i, 1])); (n*s));
%o A328248 A328248(n) = { my(k=1); while(n && !issquarefree(n), k++; n = A003415checked(n)); (!!n*k); };
%Y A328248 Cf. A003415, A008966, A256750.
%Y A328248 Cf. A328251, A005117, A328252, A328253 (indices of terms k=0, 1, 2, 3).
%K A328248 nonn
%O A328248 1,9
%A A328248 _Antti Karttunen_, Oct 11 2019