Iterated phi sequence Related: Iterated phi(n) function.
Your challenge is to compute the iterated phi function:
f(n) = number of iterations of φ for n to reach 1.

Where φ is Euler's totient function.
Related OEIS.
Here's the graph of it:


Rules:
Your goal is to output f(n) from n=2 to n=100.
This is code-golf, so shortest code wins.
Here's the values you can check against:
1, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, 3, 4, 4, 5, 3, 4, 4, 4, 4, 5, 4, 5, 4, 4, 4, 5, 4, 5, 5, 5, 5, 5, 4, 5, 4, 5, 5, 6, 4, 5, 5, 5, 5, 6, 5, 5, 5, 6, 5, 6, 4, 6, 5, 5, 5, 6, 5, 6, 5, 5, 6, 6, 5, 6, 6, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 6, 5, 6, 7, 5, 7, 5, 6, 6, 7, 5, 6, 6, 6, 6, 6, 6, 7, 5, 6, 6
Haskell, 53 52 bytes

Thanks nimi for saving 1 byte!



f&lt;$&gt;[2..100]
f 1=0
f n=1+f(sum[1|1&lt;-gcd n&lt;$&gt;[1..n]])


Try it online!

sum[1|1&lt;-gcd n&lt;$&gt;[1..n]] gives φ(n) (Taken from flawr,  thanks!)

f is a recursive function that calculates 1+φ(n) if n is not 1, and outputs 0 if n is 1, as there are no more iterations to be taken to reach 1

Finally f&lt;$&gt;[2..100] creates a list of f applied to each element of [2..100]