**Background:** The answer to a previous question I asked [here](https://mathoverflow.net/questions/224995/primitive-sequence-a-i-attaining-pillais-bound-on-sum-i-1-a-i) 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 $k=[\log \log n]$



Pruning primitive sequences but still attaining Pillai's lower bound on sum of reciprocals
Unfortunately, you have no chance whatsoever. Indeed, assume that $A$ is a subset of $\{1,\dots,n\}$ such that $[a',a'']\ge n+1$ for all $a'\ne a''$ in $A$. Then the numbers $ab: a\in A, 1\le b\le n/a$ are all different. Hence, taking the sum of their reciprocals, we obtain something like
$$
\sum\_{a\in A} a^{-1}\log(n/a)\le \log n
$$
Now, let $a\_0=ne^{-\sqrt{\log n}}$. Then $\sum\_{a\in A,a\le a\_0}a^{-1}\le \sqrt{\log n}$. On the other hand, $\sum\_{a\_0\le a\le n}a^{-1}\le\sqrt{\log n}$ as well. Thus, the trivial upper bound for the sum of the reciprocals under the new condition becomes $2\sqrt{\log n}$ or something like that.


