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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.
12
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, 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, 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
OFFSET
1,9
LINKS
FORMULA
a(4*n) = a(27*n) = 0 and in general, a(m * p^p) = 0, for any m >= 1 and any prime p.
EXAMPLE
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.
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.
PROG
(PARI)
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));
A328248(n) = { my(k=1); while(n && !issquarefree(n), k++; n = A003415checked(n)); (!!n*k); };
CROSSREFS
Cf. A328251, A005117, A328252, A328253 (indices of terms k=0, 1, 2, 3).
Sequence in context: A336387 A358217 A286133 * A360158 A329885 A085737
KEYWORD
nonn
AUTHOR
Antti Karttunen, Oct 11 2019
STATUS
approved

  NODES
orte 1
see 1
Story 1