A285320 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).

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, 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, 0, 3, 2, 1, 0, 1

Offset: 0

Author

Antti Karttunen, Apr 18 2017

Keywords

• nonn
• hard

Comments

Conjecture: all terms are well-defined (finite). This implies also the conjecture I have made in A019565.

Examples

			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.
a(74) >= 3 as A048675(74) = 2049 (squarefree), A048675(2049) =  10633823966279326983230456482242756610 (squarefree), A048675(10633823966279326983230456482242756610) = ???
		

Crossrefs

A left inverse of A109162.

Cf. A008683, A005117, A013929, A048675.

Cf. also A285319, A285331, A285332.

Programs

PARI

A048675(n) = my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; \\ Michel Marcus, Oct 10 2016
A285320(n) = if(!n || !moebius(n),0,1+A285320(A048675(n)));

Scheme

(define (A285320 n) (if (or (zero? n) (zero? (A008683 n))) 0 (+ 1 (A285320 (A048675 n)))))

Formula

If n == 0 or A008683(n) == 0, then a(n) = 0, otherwise a(n) = 1+a(A048675(n)).

a(A109162(n)) = n.