%I #16 May 19 2019 08:59:32
%S 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,
%T 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,
%U 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
%N a(n) = Euler phi function applied 5 times to n.
%H Antti Karttunen, <a href="/A049107/b049107.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A000010(A049100(n)) = A010554(A049099(n)) = phi(phi(phi(phi(phi(n))))), where phi = A000010. - _Antti Karttunen_, Aug 22 2017
%e 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.
%e 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
%t a(n)=Nest[ EulerPhi, n, 5 ]
%t Nest[EulerPhi,Range[110],5] (* _Harvey P. Dale_, May 19 2019 *)
%o (PARI) A049107(n) = eulerphi(eulerphi(eulerphi(eulerphi(eulerphi(n))))); \\ _Antti Karttunen_, Aug 22 2017
%Y Cf. A000010, A010554, A049099, A049100, A049115.
%K nonn
%O 1,41
%A _Labos Elemer_