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Rationale: Euler's totient function applied to primorials $\phi(p_n\#)=\prod_{i=1}^n(p_i-1)$ provides an exact count of the number of positive integers $\le (p_n\#-1)$ and coprime to $p_i:i\le n$. Since no positive integers $>1$ and $\le p_n$ are coprime to $p_n\#$, all enumerated coprime integers $\{r_k: 1\le k\le \phi(p_n\#)\}$ will have the form $r_k=\prod p_i^{a_i}$ for $i>n$ and $p_i<p_n\#$. Many $r_k$ will be primes, and many will be composites. I am broadly interested in the distribution of various $r_k$ into the two classes. I have chosen to focus on primorials because statements that can be shown to be true in a range bounded by consecutive primorials will be generally true, and the totient of primorials both is exactly expressible and has a clear meaning.

Framework for query: In this paragraph, I am just looking at composite $r_k$. For any composite $r_k=\prod p_i^{a_i}$, $2\le \Omega(r_k)=\sum (a_i)<n$. The upper limit of $n$ is because $p_n\#$ has $n$ prime factors and if $\Omega(r_k)\ge n$ then $r_k=\prod p_i^{a_i}$ has $n$ or more prime factors, each of which is greater than the prime factors in $p_n\#$, which would imply $r_k>p_n\#$, a contradiction. Hence $2\le \Omega(r_k)<n$. It seems reasonable to me that the upper limit on $\Omega(r_k)$ in the context of this set of conditions should be somewhat smaller than $n$. I don't see a way to find such a smaller limit, and a quick read of relevant topics on Wikipedia doesn't yield anything that addresses this point. So I turn to the community to see whether anyone has encountered this before, or can see a way proceed.

My question is: For $p_n<r_k<p_n\#\ \wedge \ r_k=\prod p_i^{a_i};\ i>n$, is there an (asymptotic?) maximum limit to $\Omega(r_k)=\sum (a_i)$ smaller than $n$? Stated in words, what is the maximum number of primes between $p_n$ and $p_n\#$ you can multiply together without the product exceeding $p_n\#$?

Keith Backman
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You meant $r = \prod_{i > n} p_i^{a_i}$. Let $\theta(x) = \sum_{p \le x} \log p= \log x\#$,

$p_n$ the $n$-th prime, $\Omega(r) = \sum_{i > n} a_i, \omega(r) = \sum_{i > n, a_i \ge 1} 1$.

Then $$\max \{ \Omega(r), r = \prod_{i > n} p_i^{a_i} \le p_n\#\} =\Omega( p_{n+1}^{\lfloor \theta(p_n)\#/\log(p_{n+1})\rfloor})$$ and $$\max \{ \omega(r), r = \prod_{i > n} p_i^{a_i} \le p_n\#\} =\max \{m-n, \theta(p_m)-\theta(p_n) \le \theta(p_n)\}$$ In both case the asymptotic is given by the PNT

$$\theta(x) \sim x, p_n \sim n \log n$$ or the Mertens theorems $$\log \theta(x) \sim \log x, \log p_n \sim \log n$$

reuns
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    I'm confused on one point: the exponent containing $\theta(p_n)#$. Based on the definition of $\theta(x)$, that would be $(log(p_n#))#$, the meaning of which isn't clear. Things don't improve if the # is moved inside the parens, as that would mean $log((p_n#)#)$. If the # is supposed to be there, could you clarify how it is interpreted? – Keith Backman Jun 07 '19 at 22:32