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

0,3

Comments

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

Links

Table of n, a(n) for n=0..73.

Formula

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

a(A109162(n)) = n.

Example

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) = ???

Prog

(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)))))

Crossrefs

A left inverse of A109162.

Cf. A008683, A005117, A013929, A048675.

Cf. also A285319, A285331, A285332.

Sequence in context: A122078 A292783 A320354 * A347710 A340666 A168068

Adjacent sequences: A285317 A285318 A285319 * A285321 A285322 A285323

Keyword

nonn,hard

Author

Antti Karttunen, Apr 18 2017

Status

approved