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In a previous mathoverflow question here a construction of a primitive sequence $1<a_1<\cdots<a_k\leq n$ formed by including all the integers in $[1,n]$ with exactly $k$ prime divisors (counted with multiplicity, so $\Omega(a_i)=k=[\log \log n]$ holds, for all $i$) was used to demonstrate how to attain Pillai's lower bound on the sum of reciprocals of primitive sequences, $$\sum_{i:a_i\leq n} \frac{1}{a_i} > c \frac{\log n}{\sqrt{\log \log n}}.$$ Clearly, not all primitive sequences necessarily attain this bound.

Now consider the primitive sequence $1<b_1<b_2<\cdots<b_l\leq n$ which is obtained by the following recipe. An integer $m\in [1,n]$ is a term in the sequence if and only if $\omega(m)=k$ for some fixed $k,$ i.e., $m$ is a product of exactly $k$ distinct primes. I expect (perhaps wrongly) that this won't make much difference, and that if not the above bound, some other nontrivial (i.e., increasing with $n$ at a reasonable growth rate) lower bound can be shown to hold for $$\sum_{i:b_i\leq n} \frac{1}{b_i}.$$

Another result proved by Behrend was on the size of the maximal $k$ for the primitive sequence of $a_i$'s. It would be interesting to know the corresponding maximal $l$ for the $b_i$'s as defined above.

My question is, how does $$\sum_{i:b_i\leq n} \frac{1}{b_i}$$ grow with increasing $n$?

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    $\begingroup$ @StefanKohl:Please see edit. $\endgroup$ – kodlu Dec 6 '15 at 14:01
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    $\begingroup$ I am sure that it has the same growth, since positive proportion of natural numbers are squarefree. $\endgroup$ – Fedor Petrov Dec 6 '15 at 14:38
  • $\begingroup$ @FedorPetrov:Thanks. Intuitively, I agree with you. However, the fact that only numbers with exactly $k$ distinct prime divisors are used to obtain the $b_i$ needs to be accounted for, no? $\endgroup$ – kodlu Dec 7 '15 at 0:03

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