Let $$f(i):=\left\lfloor\frac{p_i\#}{\varphi(p_i\#)}\right\rfloor,$$
where $p_i$ is the $i$th prime, $\#$ is the primorial operator, and $\varphi$ is totient.
Example
$$f(3)=\left\lfloor\frac{5\#}{\varphi(5\#)}\right\rfloor=\left\lfloor\frac{5\cdot 3\cdot 2}{8}\right\rfloor=3.$$
Let $g(n)$ be the smallest value of $i$ for which $f(i)=n$.
Then $g(n)\cdot \phi \approx g(n+1)$ with what seems like a high degree of accuracy. Here, $\phi=\frac{1+\sqrt{5}}{2}$.
Example
Since $f(21)=\lfloor 7.93355\rfloor$ and $f(22)=\lfloor 8.03526\rfloor$, we have $g(8)=22$. Then $22\cdot \phi\approx 35.597$, and indeed $g(9)=35$.
The $n$ values seem to roughly trace the Fibonacci sequence with some small error term. However, this may be a transient thing; see second table below, where it appears the correlation might peak around $i=F_{12}=144$. I can't calculate past $F_{20}$ using my current approach, since even that much is invoking $67931\#/\varphi(67931\#)$.
Before I dumped too much more time into investigating this, I figured I'd ask here in case this has an obvious explanation I don't see. If anyone can shed light on why this happens (or shows that my supposition is erroneous), I'll consider this answered.
Table of $(i, f(i))$
Table of $\left(a, \frac{p_{F_a}\#}{\varphi(p_{F_a}\#)}\right)$
(where $F_a$ is $a$th Fibonacci term)
I'm thinking this may be nothing more than a coincidence, per Peter's comment. In particular, I hadn't realized that you get this general sort of behavior with logs.
If you take $2 \log n$ and split it up based on integer boundaries like my $g$ function, it serves as a pretty good example of a similar quasi-recurrence-relation: