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A049108
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Number of iterations of Euler phi function needed to reach 1 starting at n (n is counted).
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15
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1, 2, 3, 3, 4, 3, 4, 4, 4, 4, 5, 4, 5, 4, 5, 5, 6, 4, 5, 5, 5, 5, 6, 5, 6, 5, 5, 5, 6, 5, 6, 6, 6, 6, 6, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 6, 7, 6, 6, 6, 7, 6, 7, 5, 7, 6, 6, 6, 7, 6, 7, 6, 6, 7, 7, 6, 7, 7, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 7, 6, 7, 8, 6, 8, 6, 7, 7, 8, 6, 7, 7, 7, 7, 7, 7, 8, 6, 7, 7, 8, 7, 8, 7, 7
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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LINKS
| R. Zumkeller, Table of n, a(n) for n = 1..10000
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FORMULA
| By the definition of a(n) we have for n >= 2 the recursion a(n) = a(Phi(n)) + 1. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 20 2001
log_3 n << a(n) << log_2 n. [Charles R Greathouse IV, Feb 07 2012]
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EXAMPLE
| If n=164 the trajectory is {164,80,32,16,8,4,2,1}. Its length is 8, thus a(164)=8.
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MATHEMATICA
| f[n_] := Length[NestWhileList[ EulerPhi, n, # != 1 &]]; Array[f, 105] (* Robert G. Wilson v, Feb 07 2012 *)
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PROG
| (PARI) a(n)=my(t=1); while(n>1, t++; n=eulerphi(n)); t \\ Charles R Greathouse IV, Feb 07 2012
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CROSSREFS
| Cf. A000010, A007755. Equals A003434 + 1.
Sequence in context: A096344 A030349 A085887 * A179846 A086925 A088858
Adjacent sequences: A049105 A049106 A049107 * A049109 A049110 A049111
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KEYWORD
| nonn,nice,easy,changed
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AUTHOR
| Labos E. (labos(AT)ana.sote.hu)
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