# Divergent sums of reciprocals with unique factorization property

Let $$A=\{a_n|n\in\mathbb{N}\}$$ be a sequence of positive integers with the following properties:

1. $$1,

2. $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\cdots=\infty$$.

Does there always exist a subsequence $$B\subseteq A$$ with $$B=\{b_n|n\in\mathbb{N}\}$$ such that

1. $$\frac{1}{b_1}+\frac{1}{b_2}+\cdots+\frac{1}{b_n}+\cdots=\infty$$,

2. All products $$d_1^{m_1}\cdot d_2^{m_2}\cdots\cdot d_k^{m_k}$$ are distinct (for all $$k\in \mathbb{N}$$), where $$d_1,\dots,d_k$$ are distinct elements of $$B$$ and $$m_1,\dots,m_k$$ are positive integers ?

• Trivial remark: the converse is not true. $1/p^2$ for p prime is a sequence with a convergent sum but which has the unique factorization property. Dec 5, 2020 at 20:48
• Is there an example when the greedy construction of $B$ (if $B\cap \{a_1,\ldots,a_{n-1}\}$ is constructed, take $a_n$ to $B$ if and only if it does not violate unique factorization) has finite sum of reciprocals? Dec 5, 2020 at 21:25
• @FedorPetrov Good question! I couldn't construct such an example. At first sight it seems that the greedy construction gives the optimal $B$. So your question might be equivalent to the problem. Dec 5, 2020 at 23:53
• Can we show that if the greedy sum converges, then so does the original sum? Dec 6, 2020 at 0:06
• @shahram ah, of course it is equivalent: the unique factorization condition means that the logarithms are independent over rationals, and the greedy algorithm finds the maximal weight independent subset (the basic property of a matroid). Dec 6, 2020 at 0:07

No. Let $$A$$ be the sequence of semiprimes $$pq$$ with $$p < q < p^2$$. Since the sum of reciprocals of primes between $$p$$ and $$p^2$$ approaches $$\ln \ln (p^2) - \ln \ln p = \ln 2$$,
$$\sum \frac1A \sim \sum_n \frac{1}{p_n} \ln 2 = \infty.$$
But any $$n$$ elements of $$B$$ whose prime factors are all at most $$p_{n - 1}$$ would violate the distinctness condition (because $$d_1^{m_1}d_2^{m_2} \dotsm d_n^{m_n} = 1$$ for $$(m_1, \dotsc, m_n) \in \mathbb Z^n$$ reduces to a linear system of $$n - 1$$ equations), so one of the first $$n$$ elements must be greater than $$\sqrt{p_n} \cdot p_n$$, and
$$\sum \frac1B < \sum_n \frac{1}{\sqrt{p_n} \cdot p_n} < \infty.$$