Primitive sequence $a_i$ attaining Pillai's bound on $\sum_{i} 1/a_i$

Question

A primitive sequence $1<a_1<\ldots<a_k\leq n$ is a sequence of integers no one of which divides any other, investigated by Erdos, Behrend and others, over the last 80 years. In fact, $\max k=\lfloor \frac{n+1}{2}\rfloor,$ which can be shown by taking exactly one member from each chain in the divisibility poset, namely $$C_x=\{x,2x,2^2x,\ldots\} \subset \{1,\ldots,n\}$$as $x$ ranges over the odd integers in $\{1,\ldots,n\}$. The easy examples of such sets, e.g., the numbers in $\{1,\ldots,n\}\cap [\lceil \frac{n}{2} \rceil,n]$ have the following property $$\sum_{i:a_i\leq n} \frac{1}{a_i} \leq c$$ for some absolute constant $c$ as $n$ increases.

However, Erdos states that S. Pillai observed that there is a constant $c'$ such that for every $n$ there is a finite primitive sequence all whose terms are upper bounded by $n$ such that $$\sum_{i:a_i\leq n} \frac{1}{a_i} > c' \frac{\log n}{\sqrt{\log \log n}}.$$

Are there any results on what such sequences may look like?

Edit: Is the value of the constant $c'$ or bounds on it known?

Answer 1

Fix $k$ and consider the set $E_k$ of all numbers with exactly $k$ prime factors (counted with multiplicity). They definitely form a primitive sequence. Already for $k=1$ sum of reciprocals is unbounded. I think, for $k=[\log \log n]$ we should obtain lower bound like $\log n/\sqrt{\log\log n}$. It is natural to expect: randomly chosen number in $[1,n]$ has exactly $k$ prime factors with probability about $1/\sqrt{\log\log n}$ by Erdős-Kac theorem

Denote $k=[\log \log n]$. Then by result of Selberg cited by alpoge in the comments, $|E(k)\cap [1,x]|\geq C x/\sqrt{k}$ for all $x\in [\sqrt{n},n]$. It follows that $$ \sum_{m\in E(k)\cap [1,n]} m^{-1}\geq \sum_{m\in E(k),n\geq v\geq m} \frac1{v(v+1)}\geq \sum_{v=\sqrt{n}}^n \frac1{v(v+1)}|E(k)\cap [1,v]|\geq c\log(n)/\sqrt{k} $$ for some $c>0$.