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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). 7
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
Conjecture: all terms are well-defined (finite). This implies also the conjecture I have made in A019565.
LINKS
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. also A285319, A285331, A285332.
Sequence in context: A122078 A292783 A320354 * A347710 A340666 A168068
KEYWORD
nonn,hard
AUTHOR
Antti Karttunen, Apr 18 2017
STATUS
approved

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Last modified April 25 10:47 EDT 2024. Contains 371967 sequences. (Running on oeis4.)