Let , where denotes the Euler's totient function . Find the minimum value of such that .
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ϕ n denotes how many times Euler's totient function can be applied before the result reaches 0 - an iterative form of Euler's totient function.
From this:
ϕ ( 1 8 0 ) = 4 8 ( 1 )
ϕ ( 4 8 ) = 1 6 ( 2 )
ϕ ( 1 6 ) = 8 ( 3 )
ϕ ( 8 ) = 4 ( 4 )
ϕ ( 4 ) = 2 ( 5 )
ϕ ( 2 ) = 1 ( 6 )
Therefore, n = 6