We can write any integer in the form $m^h r$, where m is square-free, and $r$ is h-free and coprime to $m$.
For example, $ \ 10584 = 2^3\cdot 3^3 \cdot 5\cdot 7^2 = 6^3 \cdot 245$
This way, we can generate the values of k, already knowing the values of m and h. Also, note that $H_k$ is equal to $h$, and $b_k$ and $B_k$ are the smallest and largest prime factors of $m$ respectively.
Using this knowledge,
$$\sum_{k = 2}^n{\frac{b_k}{B_k}}\tag{1}$$
can be re-written as
$$\sum_{h\ge 2}{\sum_{r\le n}{q_h(r)\sum_{m^hr\le n, \ m \ge 2}{\left(q_2(m)\cdot[gcd(m, r)= 1]\cdot\frac{\text{smallest prime factor of m}}{\text{largest prime factor of m}}\right)}}},\tag{2}\\ where \ \ q_h(r) = \begin{cases}
1: & \text{if $r$ is h-free} \\
0: & \text{otherwise}
\end{cases}.$$
Now it just remains to find some asymptotic for
$$\sum_{m = 2}^x{\left(q_2(m)\cdot[gcd(m, r) = 1]\cdot\frac{\text{smallest prime factor of m}}{\text{largest prime factor of m}}\right)}\tag{3}$$
in terms of x.
We can re-use some of the results and techniques from this source - https://arxiv.org/pdf/1907.09129.pdf
By writing the smallest and largest prime factors of $m$ as $p(m)$ and $P(m)$ respectively, we have
$$\sum_{m = 2}^x{\left(q_2(m)\cdot[gcd(m, r) = 1]\cdot\frac{p(m)}{P(m)}\right)} \ = \ \sum_i{\sum_{2 \le m \le x \\ \omega(m) = i}{\left(q_2(m)\cdot[gcd(m, r) = 1]\cdot\frac{p(m)}{P(m)}\right)}}\\$$
$$=\sum_{p \le x}{\left(q_2(p)\cdot [gcd(p, r) = 1] \cdot \frac{p}{p}\right)} \ + \ O\left(\sum_{i \ge 2}{\sum_{2 \le m \le x \\ \omega(m) = i}{\frac{p(m)}{P(m)}}}\right)\\ = \ \sum_{p \le x \\ p \not\mid r}{1} \ + \ O\left(\frac{x}{\log^2(x)}\right)\tag{4}$$
where the bounding of the RHS came from the source above.
Clearly, the main term in the limit comes from the summation on the left, thus we need only plug this back into $(2)$, with $x = (\frac{n}{r})^{\frac{1}{h}}$.
This gives,
$$\sum_{h\ge 2}{\sum_{r\le n}{q_h(r)\sum_{p \le (\frac{n}{r})^{\frac{1}{h}} \\ p \not\mid r}{1}}} \ = \ \sum_{h\ge 2}{ \ \sum_{ \ p^h \le n}{ \ \sum_{r \le \frac{n}{p^h} \\ p \not\mid r}{q_h(r)}}}\tag{5}$$
Now, using the fact that $q_h(r) = \sum_{d^h \mid r}{\mu(d)}$, one can find that
$$\sum_{r \le y \\ p \not\mid r}{q_h(r)} \ \sim \ \frac{1}{\zeta(h)} \cdot \frac{p^h - p^{h-1}}{p^h - 1} \cdot y \tag{6}$$
And substituting into $(5)$ leads to our final answer
$$\sum_{h\ge 2}{ \ \sum_{ \ p^h \le n}{\frac{1}{\zeta(h)} \cdot \frac{p^h - p^{h-1}}{p^h - 1} \cdot \frac{n}{p^h}}}\tag{7}$$
The $n$ cancels with the $\frac{1}{n}$ in the limit, hence the final answer is
$$c = \sum_{h\ge 2}{\frac{1}{\zeta(h)} \sum_{p}{\frac{p - 1}{p(p^h - 1)}}}. \tag{8}$$
Edit:
I got a value of $0.36604$ as a close upper bound to this sum.
The comments are getting kinda long, so here it is.
I checked the larger database thanks to @MartinHopf.
I checked the database by checking for each pseudoprime if it is a Carmichael number and then finding the greatest prime factor of the number, I eliminated $p$ as possible addition to $S$ if I found that $p$ was the greatest factor of such a number.
The remaining candidates below $50\ 000$ are [25799, 40883, 42023, 42443, 42683, 42767, 44159].
Update:
77579217225595049581 is an example for $p = 25\ 799$.
The remaining candidates below $50\ 000$ are now [40883, 42023, 42443, 42683, 42767, 44159].
Update:
192231276682846353121 and 30691624706028820801 are examples for $p=40\ 883$.
361362050102075568481 for $p = 42\ 023$.
187245600667639628401 for $p = 42\ 443$.
1421298370313927469001 and 1475526798515376902401 for $p = 42\ 683$.
3276241962960743496721 and 824592027949741719361 for $p = 42\ 767$.
And
18474317205530955301
25086959880820862401
32793227139330403201
35249365392461726881
626067460857971232001
20726926900955087905
941595098724511030801
1947595024403123504401
7625997271344313144921
1191134505408294459601
18598551907931088038401
3290982924650209776001
17875561355805607977121
11058363534498776939521
2426379943337261168401
are all examples for $p=44\ 159$.
There are no remaining candidates below $50\ 000$.
Update: I raised the search limit to all $p$ below $100\ 000$ by eliminating the $120$ possible candidates (derived from the dataset) for $p$ between $50\ 000$ and $100\ 000$.
The algorithm I used to eliminate the candidates rests on finding primes $q$ and $r$ smaller than $p$ such that $n=mpqr$ (where $m$ is varied to search multiple numbers) is a Carmichael number with maximum factor $p$. (I first only used $p$ and $q$ but found that using three factors is even faster). Since $n$ is Carmichael, we have $p-1|n-1$, $q-1|n-1$ and $r-1|n-1$. Thus $n \equiv 1 \equiv mpr \pmod {q-1}$, $mqr \equiv 1 \pmod{p-1}$ and $mpq \equiv 1 \pmod{r-1}$. Using this we can find (using the Chinese Remainder Theorem) necessary congruences for $m$ and speed up our search.
Best Answer
Conjecturally (on the Hardy–Littlewood prime 3-tuples conjecture) the answer is yes.
The following is shown in Chapter 14 of O. Ore's Number Theory and Its History and is not too hard to verify by hand: Given any three positive, pairwise relatively prime integers $a<b<c$, let $m$ be the solution to the congruence $m(ab+bc+ca)+a+b+c\equiv0\pmod{abc}$. Then for any integer $n \equiv m\pmod{abc}$, if each of $an+1$, $bn+1$, and $cn+1$ is prime, then $(an+1)(bn+1)(cn+1)$ is a Carmichael number.
Conjecturally that simultaneous primality should happen infinitely often; and in this family the ratio of the largest to smallest prime factor approaches $c/a$ as $n\to\infty$, which can be chosen arbitrarily close to $1$ (by choosing $(a,b,c)=(2d-1,2d,2d+1)$ for example).