Does the mean ratio of the bases of the largest prime factor exponents converge? Let $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ and $H_n = \max(a_1, a_2, \ldots, a_k)$ be the largest exponent in the prime factorization of $n$. It is possible that two or more prime factors can both be the bases of the largest exponent $H_n$. Let $b_n$ and $B_n$ be the smallest and the largest prime factor of $n$ respectively for which the exponent is $H_n$. Clearly, if there is only one distinct prime which have the largest exponent then $b_n = B_n$ otherwise $b_n < B_n$.
Example: If $n = 2^3 5^2$ then $b_n = B_n = 2$ and if $n = 2^2 3^2 5$ then $b_n = 2$ while $B_n = 3$.
Experimental data shows that the mean value of the ratios $\frac{b_n}{B_n}$ converges to some value $0.392$ and $0.45$. Below are values of the mean in two intervals of different orders of magnitude.

*

*$0.39190 < c < 0.45$ in $(2,3.5 \times 10^9)$
Also note that the limiting value is suspiciously close to $\frac{\pi}{8} \approx 0.3926$ though this might be a coincidence.

Conjecture: There is a constant $0.392 < c < 0.45$ such that,
$$ \lim_{n \to \infty} \frac{1}{n}\sum_{k = 1}^n \frac{b_k}{B_k} = c $$

Can this be proved or disproved?
SageMath Code:
n = 2
sum_base_min = 0
sum_base_max = 0
sum_ratio = 0
sum_ratio_2 = 0
step = 10^7
target = step

while True:
    f = prime_factors(n)
    prime_power = []

for factor in f:
    a = 1
    check = 1
    while True:
        check = check*factor
        if n%check == 0:
            a = a + 1
        else:
            a = a - 1
            t = (factor,a)
            prime_power = prime_power + [t]
            break

max_power = 1
min_power = n
min_base  = 2

for item in prime_power:
    power = item[1]
    if power > max_power:
        max_power = power
        min_base = item[0]

# To get the lower bound, set an alternate minimum base to 2 instead of
# the smallest prime factor for square free numbers 

if max_power == 1:
    min_base = prime_power[0][0]
    min_base_2 = 2 
else:
    min_base_2 = min_base

max_base = min_base

for item in reversed(prime_power):
    if power == max_power:
        max_base = item[0]
        break
        
sum_base_min = sum_base_min + min_base
sum_base_max = sum_base_max + max_base

sum_ratio   = sum_ratio + min_base/max_base.n()
sum_ratio_2 = sum_ratio_2 + min_base_2/max_base.n()

# print(n, prime_power, max_power, min_base_2, min_base, max_base)

if n == target:
    print(n, sum_base_min/n.n(),sum_base_max/n.n(), sum_ratio_2/n, sum_ratio/n)
    target = target + step

n = n + 1

 A: 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.
