Is there a smallest $r$ such that $n+\varphi(n)=\displaystyle \prod_{i=1}^r q_i$ always has solutions for mutually different odd primes $q_i $? While discussing with Peter in one of the chatrooms on MSE I proposed an idea to try to find smallest natural number $r$ such that $n+\varphi(n)=\displaystyle \prod_{i=1}^r q_i$ has solutions for every choice of mutually different $r$ odd primes $q_i:i=1,...,r$.
Of course, I do not know is there really such an $r$, therefore this question.
Because $n+ \varphi(n)$ is "relatively dense" in $\mathbb N$ it seems that there could be such a smallest $r$ since as $r$ becomes larger and larger the product $\displaystyle \prod_{i=1}^rq_i$ becomes "nicely" dense (but how much and how much nicely) in $\mathbb N$, so it seems, at least from a naive standpoint, that some $r$ could strike a sensible balance between the two requirements.
But then, "on the other hand", the requirement that there is a solution for every choice of $r$ mutually different odd primes seems as to really be "too much" to be fulfilled. 
Peter computed the smallest numbers for which there is no $n$ that is a solution for $r=1,2,3,4,5,6,7,8$ and these are:
$7, 35, 195, 1155, 15015, 255255, 4849845, 140645505$
This sequence is not in OEIS.
Furthermore,  if $2$ is allowed to be one of the primes, which we do not consider in this question, then the corresponding sequence is also not in OEIS. 
The question:


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*Is there a smallest natural number $r$ such that $n+\varphi(n)=\displaystyle \prod_{i=1}^r q_i$ has solutions for every choice of mutually different $r$ odd primes $q_i:i=1,...,r$?


I can admit I didn´t think too much about this problem, but straightforwardly typed it here because I know that some folks here know about these topics more than I, so I can expect an answer and(or) a comment which settles the "whole thing".
Now, when I again think of this, for small $r$ this product is rather "wildly" distributed over $\mathbb N$, but also is for large $r$.
So some density arguments seem to be surely needed, it would be next to a wonder if $n + \varphi(n)$ could really "cover" all the products for some $r$, in whose existence I seriously doubt.
 A: This is just an idea may it help you. I think your problem related to the primorial numbers. In particular rate growth of $\frac{\phi(n)}{n}$, Nicolas showed that there are infinitely many numbers, primorials, such that:
$$  \frac{n}{e^\gamma \log \log n + \frac{2.50637}{\log \log n}} < \varphi(n) < \frac{n}{e^\gamma \log \log n  }.$$ You may check here the paper of Nicolas, in French. Then, we assume the existence of infinity of premorial numbers. This means that:$$  \frac{1}{e^\gamma \log \log n + \frac{2.50637}{\log \log n}} <\frac{ \varphi(n)}{n} < \frac{1}{e^\gamma \log \log n  }.$$ Now you can replace $\frac{\varphi(n)}{n}$  by $\displaystyle \frac{\prod_{i=1}^r q_i}{n}-1 $ we get:
$$  \frac{1}{e^\gamma \log \log n + \frac{2.50637}{\log \log n}} <\frac{\prod_{i=1}^r q_i}{n}-1 < \frac{1}{e^\gamma \log \log n  }.$$
Really I suspect there is such an $r$ using the latter identity but really I don't know the way to get it only we may look to when does that product in the last identity close to $1$ because squeeze theorem would be hold in this case.
ADDENDUM Also another question  we may ask here is what is the smallest $r$ for which $n\mid(\prod_{i=1}^r q_i)$ in the last identity? But this seems so hard.
