TITLE: Can it be proven/disproven that there are highly composite numbers that prime-factorize into larger primes such as $9999991$?
QUESTION [22 upvotes]: Of course, following the rules found by Ramanujan, such a highly composite number would need to factorize into all primes ascending up to 9999991 (with descending powers as the primes progress) so the highly composite number would be insanely large.
However a highly composite number needs more factors than all other numbers before it, so surely any number with very large prime factors like 9999991 is automatically at a disadvantage?
So is there a limit to the size of the largest prime factor of a highly composite number, or is it limitless? Is there even a way to know?

REPLY [13 votes]: Given a prime $p,$ there is always a highly composite number which has $p$ as a factor. In fact, there is always a highly composite number for which $p$ is the largest prime factor.
There is a subsequence that Ramanujan called the Superior Highly Composite Numbers. These numbers are always highly composite. Given a real $\delta > 0,$ we construct 
$$  N_\delta = \prod_{i=1}^\infty  p_i^{a_i},$$
where only a finite number of the exponents $a_i$ are nonzero, with specific value
$$  a_i = \left\lfloor \frac{1}{p_i^\delta - 1} \right\rfloor. $$
The smallest SHC number which is divisible by prime $9999991$
uses
$$  \delta = \frac{\log 2}{ \log 9999991} \approx 0.043004. $$
I calculate that the exponent of the prime $2$ is $33,$ exponent of the prime $3$ is $20.$ Go Figure.
Let's see, the number $N_\delta$ is, very roughly, $e^{9999991} \approx 10^{4342941}$