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%I #16 Apr 18 2017 15:43:27
%S 0,1,2,3,0,1,4,1,0,0,2,1,0,1,1,5,0,1,0,1,0,3,2,1,0,0,2,0,0,1,2,1,0,1,
%T 2,1,0,1,3,3,0,1,2,1,0,0,2,1,0,0,0,3,0,1,0,1,0,3,1,1,0,1,1,0,0,1,2,1,
%U 0,3,2,1,0,1
%N If n == 0 or A008683(n) == 0, then a(n) = 0, otherwise a(n) = 1+a(A048675(n)); number of iterations of A048675 needed before the result is either zero or nonsquarefree number (A013929).
%C Conjecture: all terms are well-defined (finite). This implies also the conjecture I have made in A019565.
%F If n == 0 or A008683(n) == 0, then a(n) = 0, otherwise a(n) = 1+a(A048675(n)).
%F a(A109162(n)) = n.
%e a(38) = 3 because 38 = 2*19 (thus squarefree), A048675(38) = 129 (= 3*43), A048675(129) = 8194 (= 2*17*241) and A048675(8194) = 4503599627370561 (= 3^2 * 37 * 71 * 190483425427), so three steps were needed before nonsquarefree number was reached.
%e a(74) >= 3 as A048675(74) = 2049 (squarefree), A048675(2049) = 10633823966279326983230456482242756610 (squarefree), A048675(10633823966279326983230456482242756610) = ???
%o (PARI)
%o A048675(n) = my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; \\ _Michel Marcus_, Oct 10 2016
%o A285320(n) = if(!n || !moebius(n),0,1+A285320(A048675(n)));
%o (Scheme) (define (A285320 n) (if (or (zero? n) (zero? (A008683 n))) 0 (+ 1 (A285320 (A048675 n)))))
%Y A left inverse of A109162.
%Y Cf. A008683, A005117, A013929, A048675.
%Y Cf. also A285319, A285331, A285332.
%K nonn,hard
%O 0,3
%A _Antti Karttunen_, Apr 18 2017