For reasons of personal curiosity, I was playing with the sum of iterates of $\phi (n)$, which are listed in OEIS A053478; that is $a(n)=n+\phi (n)+\phi(\phi (n))\dots \ $ The sum terminates when the final term becomes $1$.
Since the sum contains $n$ explicitly, $\frac{a(n)}{n} \ge 1$, and in fact is $1$ only at $n=1$
However, by direct observation of the first several hundred terms, I find that $\frac{a(n)}{n}<3$. The ratio closely approaches $3$ only for some prime numbers. I know that when $n$ is prime, $a(n)>n+(n-1)\approx 2n$. However, despite much cogitation, I cannot reach a concise understanding of why the superior limit of $a(n)$ appears to be $3n$, and so I cannot establish that $3n$ is in fact the limit. The possibility remains open that there may exist some very large $n$ for which $\frac{a(n)}{n}>3$.
My question is: Can it be proved that the superior limit of $\frac{a(n)}{n}$ is in fact $3$?
It is true that $\frac{a(n)}{n}<3$. We know
$(1)$ $ \phi(n)\leq \frac{n+1}{2}$ if $n$ is even,
$(2)$ $ \phi(n)\leq n-1$ always,
$(3)$ $\phi(n)$ is always even,
Thus we have that
$n+\phi(n)+\phi(\phi(n))+\ldots \leq n+n-1+\sum_{k=1}^{\log_2(n-1+1)} \frac{n}{2^k}=2n-1+n-1=3n-2$,
as desired. Also this yields that $\lim \sup \frac{a_n}{n}\leq 3$.
It makes sense that the prime cases are the closest to $3n$ because there is only one term in the sum that is odd (the first term $n$) and prime $n$ maximize the next term to be $n-1$ (every successive term basically decreases by a factor of $2$ at least).
Edit: If there are infinitely many Fermat primes $F=2^{2^{j}}+1$ (unlikely), then we have that
$a(F)=F+\sum_{k=0}^{2^j} 2^k> 3F-3$, infinitely often, in which case
$\lim \sup \frac{a(n)}{n} = 3$.
If there are not infinitely many Fermat primes, I'm not sure if the superior limit is $3$ or less than $3$.
Edit 2: By Linnik's theorem and the GRH (we can have a slightly worse estimate w/o GRH) we know that in an arithmetic progression like $N=J2^k+1$, $J=1,2,…$, there is a prime $N\leq \phi^2(2^k)\ln^2(2^k)< k^2\cdot 2^{2k-2}$ (Wikipedia link). In particular there is a prime in that sequence for $J<k^22^{k-2}\leq (\log(N)+2)\sqrt{N}$. Thus we have that infinitely many $N$ (there are infinitely many $k$ to choose; also the below calculation is true for $N\geq 35$) such that
$a(N)\geq N+N-1+\sum_{k=1}^{\log_2(\frac{N-1}{(\log(N)+2)\sqrt{N}})} \frac{N-1}{(\log(N)+2)\sqrt{N}\cdot 2^k}> 2N+N\frac{\sqrt{N}}{2N\log(N)}$.
This is not good enough to get a better lower bound than $\lim \sup \frac{a_n}{n}\geq 2$, which we already knew. If we had some fixed constant $K>1$ such that there are infinitely many $k$ such that the sequence $J2^k+1$, $J=1,2,...$ will have a prime $J\leq K$, then
$\lim \sup \frac{a_n}{n}\geq 2+1/K$
by the same calculation as above.
Conjecture: $2.75\geq \lim \sup \frac{a_n}{n}\geq 2.5$.