TITLE: Is there an integer for every Euler rank?
QUESTION [10 upvotes]: Let $φ$ be the Euler function, then $φ(n)+1 = n$ iff $n$ is prime.
Let the map $\psi: n \mapsto φ(n)+1$  
An integer $m$ is of Euler rank $0$ if it is prime, and of Euler rank $r$ if $\psi^{(r)}(m)$ is prime whereas $\psi^{(r-1)}(m)$ is not. The following list shows the smallest integer $n \ge 2$ of rank $r \le 15$:  
$\scriptsize{ \begin{array}{c|c}
 r  &0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15 \newline 
                             \hline
n &2&4&15&35&69&255&535&949&1957&2513&2923&4531&17701&22957&54589&79421  
\end{array} }  $  
Question: Let $r \ge 0$,  is there an integer $n \ge 2$ of  Euler rank $r$?

REPLY [5 votes]: Not an answer, but an extension of the sequence with PARI/GP
? for(j=0,40,m=2;s=0;while(s<>j,m=m+1;n=m;s=0;while(isprime(n)==0,s=s+1;n=eulerp
hi(n)+1));print(m," ",s))
2 0
4 1
15 2
35 3
69 4
255 5
535 6
949 7
1957 8
2513 9
2923 10
4531 11
17701 12
22957 13
54589 14
79421 15
80029 16
84493 17
98581 18
102827 19
115243 20
239111 21
291149 22
310813 23
362621 24
398893 25
598341 26
801923 27
838307 28
1063493 29
1079833 30
1123813 31
1311121 32
1329403 33
1582439 34
1677931 35
1751831 36
2382469 37
7754613 38
70817623 39
94342651 40

The sequence of the smallest numbers with a given rank must strictly increase because for
every composite number $n$, we have $\phi(n)+1<n$. If the smallest number with
rank $r$ is $a$ and the smallest number with rank $r+1$ is $b$, then the assumption $a\ge b$ together with rank $(\phi(b)+1)=r$ and $\phi(b)+1<b\le a$ contradicts the choice of $a$.