A049100 a(n) = Euler phi function applied 4 times to n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 4, 1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 1, 4, 2, 2, 2, 4, 2, 4, 2, 2, 4, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 2, 4, 8, 2, 8, 2, 4, 4, 8, 2, 4, 4, 4, 4, 4, 4, 8, 2, 4, 4, 8, 4, 8, 4, 4

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Examples

			n=163, the successive iterates applying Euler totient function are as follows: 163,162,54,18,6,2,1. The 5th term is 6, when Phi was applied 4 times. So a(163)=6.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Boris Putievskiy, Transformations [Of] Integer Sequences And Pairing Functions, arXiv preprint arXiv:1212.2732 [math.CO], 2012.

Crossrefs

Cf. A000010, A010554, A049099, A049107, A049115.

Programs

Maple

with(numtheory): seq(phi(phi(phi(phi(n)))),n=1..130); # Emeric Deutsch, May 14 2006

Mathematica

a(n)=Nest[ EulerPhi, n, 4 ]

PARI

A049100(n) = eulerphi(eulerphi(eulerphi(eulerphi(n)))); \\ Antti Karttunen, Aug 22 2017

Formula

a(n) = phi(phi(phi(phi(n)))) = A000010(A000010(A000010(A000010(n)))) = A010554(A010554(n)) = A000010(A049099(n)).

Extensions

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 23 2007