What does the sum of the reciprocals of all the highly composite numbers converge to?

Question

I've calculated the sum of the reciprocals of all the $156$ first highly composite numbers up to $10^{18}$:

$\sum \dfrac{1}{HCC(n)} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dfrac{1}{36} + ... \approx 1.1328728$

It seems to converge to this number... but is there a mathematical formula which spawns this number? Or is this number completely random?

I've noticed that it's close to $e^{\frac{1}{8}} \approx 1.133148$ but this may just be a coincidence.

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I've also calculated the sum of the reciprocals of all the $69$ first superior highly composite numbers up to $10^{95}$:

$\sum \dfrac{1}{SHCC(n)} = \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{60} + \dfrac{1}{120} + ... \approx 0.77839341508236$

So again same question: does this number have a mathematical formula which spawns it?

I've noticed that it's close to $e^{-\frac{1}{4}} \approx 0.7788008$ but again it's also maybe a coincidence.

Answer 1

Just some thoughts I wanted to share. I guess they are random numbers, which we can not express in terms of other popular constants.

However, let $Q(n)$ be the number of highly composite numbers $\leq n$. Then it is known that $(\ln(x))^a\leq Q(x)\leq(\ln(x))^b$ for some $a$ and $b$ bigger than 1. Assuming $Q(x)\approx (\ln(x))^c $ for some $c>1$ and interpreting $Q'(x)$ as the density of highly composite numbers, we can naively write \begin{equation} \sum\frac{1}{HCC(n)}\approx \int_{x=1}^\infty Q'(x)\frac{1}{x}\text{ d}x\approx \int_{x=1}^\infty \frac{c(\ln (x))^{c-1}}{x^2}\text{ d}x=\Gamma[c+1].\end{equation}

Now, here http://wwwhomes.uni-bielefeld.de/achim/highly.html it is said that $c\approx \frac{5}{4}$. Finally, $\Gamma[\frac{5}{4}+1]\approx 1.133$, which is close to your numerical value.