**Background:** The answer to a previous question I asked here specified a construction to achieve Pillai's bound on reciprocal sums of primitive sequences. A primitive sequence $1<a_1<\ldots<a_k\leq n$ is a sequence of integers no one of which divides any other. Pillai proved that
for every $n$ there is a finite primitive sequence all whose terms are upper bounded by $n$ such that
$$\sum_{i:a_i\leq n} \frac{1}{a_i} > c' \frac{\log n}{\sqrt{\log \log n}}\quad(1).$$
The construction that achieves this bound is to take all numbers with exactly $k=[\log \log n]$ (not necessarily distinct) prime divisors. Call this set of numbers $A_n.$

**Goal:** We would like to remove `bad`

terms from $A_n$ such that for the remaining subsequence $\{a_{i_s}\}$ the following holds:$$[a_{i_s},a_{i_s'}]\geq n+1,$$ whenever $i_s\neq i'_s$.Here, $[x,y]$ denotes the least common multiple of integers $x,y.$

The hope is that there aren't too many bad terms, i.e., that the remaining terms, when their reciprocals are summed, still satisfy a lower bound not much smaller than (1).

Consider $k=2$ for clarity. Note that the smallest LCM occurs if there is a common prime in the two numbers thus, $[pq,p'q']=[p'q,p'q']\geq p'qq',$ for example. If we remove $a_i$ which are divisible by primes $\leq n^{1/(k+1)}$ then our condition (1) is satisfied. Now compute the sum over bad $a_i$ that we need to subtract form the LHS of (1): $$B_2=\sum_{p\leq n^{1/3}} \frac{1}{p} \sum_{p\leq q\leq \sqrt{n/p}} \frac{1}{q}+ \sum_{p,q\leq n^{1/3}} \frac{1}{pq}\quad(2).$$ The second term in (2) is $(\log\log (n^{1/3}))^2=(\log \log n -\log 3)^2\leq (\log \log n)^2$ and is negligible compared to the RHS of (1).

The first term can be rewritten as $$\sum_{p\leq n^{1/3}} \frac{1}{p} \left[ \log\log(\sqrt{n/p})-\log\log p \right] $$ and after some manipulation becomes (if I'm right) $$ \log(\frac{1}{2}) \sum_{p\leq n^{1/3}} \frac{1}{p} +\log\log n\sum_{p\leq n^{1/3}} \frac{1}{p}-\frac{1}{\log n} \sum_{p\leq n^{1/3}} \frac{\log p}{p} - \sum_{p\leq n^{1/3}} \frac{\log \log p}{p}. $$ which turns out to be $\ll (\log\log n)^2.$ Thus $B_2 \ll (\log\log n)^2.$

Is this analysis correct? More importantly, as $n$ increases, so does $k$ and it seems hard to extend this analysis for $B_k$. Is there a more general and powerful result that can be used to conclude that perhaps $$B_k \ll (\log\log n)^{c}$$ for some absolute constant $c$?

In any case, the main goal is for $B_k$ to be shown to be strictly smaller than the RHS of (1) for $k=[\log \log n]$