A049107 a(n) = Euler phi function applied 5 times to n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 4, 1, 4, 1, 2, 2, 4, 1, 2, 2, 2, 2, 2, 2, 4, 1, 2, 2, 4, 2, 4, 2, 2

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Examples

			For n = 163, the successive iterates applying Euler totient function are as follows: 163, 162, 54, 18, 6, 2, 1. The 6th term is 2, when Phi was applied 5 times. So a(163)=2, already a power of 2.
For n = 487, the successive iterates are 486, 162, 54, 18, 6, 2, 1. On the fifth iteration we reach 6, thus a(487) = 6. This is also the first term of A049107 that is not a power of 2. - _Antti Karttunen_, Aug 22 2017
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000010, A010554, A049099, A049100, A049115.

Programs

Mathematica

a(n)=Nest[ EulerPhi, n, 5 ]
Nest[EulerPhi,Range[110],5] (* Harvey P. Dale, May 19 2019 *)

PARI

A049107(n) = eulerphi(eulerphi(eulerphi(eulerphi(eulerphi(n))))); \\ Antti Karttunen, Aug 22 2017

Formula

a(n) = A000010(A049100(n)) = A010554(A049099(n)) = phi(phi(phi(phi(phi(n))))), where phi = A000010. - Antti Karttunen, Aug 22 2017