---

Statement

Let

$$\lambda(N)=\max\left\{\sum_{n\in A}\frac1n: A\subseteq\{1,\ldots,N\},\ \text{no }S\subseteq A\text{ has }\sum_{n\in S}\frac1n=1\right\}.$$

Then $\lambda(N)\gg(\log\log N)^2$.

Source. Bloom, arXiv:2112.03726v2, Theorem 4, Appendix A, p. 19. Bloom attributes the construction to Pomerance, via personal communication reported in Croot's 2000 thesis. This is a distinct obstruction method, not an alternative proof of the positive-density theorem.

Rewritten proof

Fix a large absolute constant $C$. Let $A$ consist of those $n\le N$ whose largest prime divisor $p$ satisfies $p\log p>Cn$. For every sufficiently large prime $p\le N/\log N$, all $n=pm$ with $1\le m<\log p/C$ lie in $A$: here $m<p$, so $p$ is the largest prime factor, and $pm<N$. Different choices of largest prime produce disjoint sets. Hence

$$R(A)\ge\sum_{p\le N/\log N}\frac1p \sum_{1\le m<\log p/C}\frac1m \gg\sum_{p\le N/\log N}\frac{\log\log p}{p} \gg(\log\log N)^2.$$

The finitely many small primes can be omitted. For the last estimate, partial summation of Mertens' $\sum_{p\le t}1/p=\log\log t+O(1)$ gives $\sum_{p\le X}(\log\log p)/p =\frac12(\log\log X)^2+O(\log\log X)$.

Suppose distinct $n_1,\ldots,n_k\in A$ have reciprocals summing to one. Choose the largest prime $p$ dividing any denominator, and label those divisible by $p$ as $pm_1<\cdots<pm_r$. For each such term, $p$ is its largest prime divisor, so $m_j<\log p/C<p$. In particular none of the $m_j$ is divisible by $p$. We have

$$\frac1p\sum_{j=1}^r\frac1{m_j} =1-\sum_{j=r+1}^k\frac1{n_j},$$

whose right side has reduced denominator coprime to $p$. If $D=\operatorname{lcm}(m_1,\ldots,m_r)$ and $T=D\sum_j1/m_j$, then $T$ is a positive integer and $p\nmid D$. The denominator condition forces $p\mid T$, whence $p\le T$. Put $m=m_r$. Then

$$p\le T\le\operatorname{lcm}(1,\ldots,m)\sum_{j=1}^m\frac1j \le e^{C_0m}$$

for an absolute $C_0$, using Chebyshev's estimate $\log\operatorname{lcm}(1,\ldots,m)=O(m)$ and absorbing the harmonic factor. Thus $\log p\le C_0m$, contradicting $\log p>Cm$ if $C>C_0$. Therefore $A$ has no unit subsum, proving the bound.

Dependencies and method

External Mertens and Chebyshev estimates. The transferable mechanism is to isolate the largest prime among a proposed representation and bound the positive integer it must divide. The complementary upper bound in this source is Theorem 3. No claim is made here that these are the best bounds in all subsequent literature.

Bears on

• Problem 47 (the construction shows that Erdős's speculated threshold $(\log\log N)^2$ would be best possible)
• Problem 298
• Problem 299