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A073407
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Let phi_m(x) denote the Euler totient function applied m times to x. Sequence gives the minimum number of iterations m such that phi_m(n) divides n.
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0
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1, 1, 2, 1, 3, 1, 3, 1, 3, 2, 4, 1, 4, 2, 4, 1, 5, 1, 4, 2, 4, 3, 5, 1, 5, 3, 4, 2, 5, 3, 5, 1, 5, 4, 5, 1, 5, 3, 5, 2, 6, 3, 5, 3, 5, 4, 6, 1, 5, 4, 6, 3, 6, 1, 6, 2, 5, 4, 6, 3, 6, 4, 5, 1, 6, 4, 6, 4, 6, 4, 6, 1, 6, 4, 6, 3, 6, 4, 6, 2, 5, 5, 7, 3, 7, 4, 6, 3, 7, 4, 6, 4, 6, 5, 6, 1, 7, 4, 6, 4, 7, 5, 7, 3, 6
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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FORMULA
| It seems that sum(k=1, n, a(k)) is asymptotic to C*n*log(n) with C>1.
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EXAMPLE
| phi(22) ->10 phi(10)->4 phi(4)->2 and 2 divides 22. Hence 3 iterations are needed and a(22) = 3
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PROG
| (PARI) a(n) = if(n<0, 0, c=1; s=n; while(n%eulerphi(s)>0, s=eulerphi(s); c++); c)
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CROSSREFS
| Cf. A019294.
Sequence in context: A152188 A025820 A109704 * A178810 A049994 A135732
Adjacent sequences: A073404 A073405 A073406 * A073408 A073409 A073410
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KEYWORD
| easy,nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 23 2002
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