# Size of finite subset of $\mathbb{N}$ such that the sum of reciprocals is a given positive integer

Let $$\mathbb{N}$$ denote the set of positive integers. For every integer $$k\in\mathbb{N}$$ let $$m(k)$$ denote the minimal size of a finite set $$S\subseteq \mathbb{N}$$ such that $$\sum_{j\in S}j^{-1}=k$$.

What is the asymptotic growth of $$m(k)$$?

• I think this is somewhat related to harmonic numbers and the euler constant – vidyarthi May 27 '19 at 9:58
• The sum of $N$ different Egyptian fractions (fractions with numerator 1 and positive integer denominator) is at most $\log N+O(1)$, that implies the exponential lower bound for $m(k)$. – Fedor Petrov May 27 '19 at 10:16
• The keyword for this is "Egyptian fractions". A related question is at math.se ... math.stackexchange.com/q/3185675/442 ... no "answer" is given there, either. – Gerald Edgar May 27 '19 at 10:51
• ... but a comment there does mention the exponential lower bound. – Gerald Edgar May 27 '19 at 11:47
• @DominicvanderZypen I suggest to change the question to "what is the asymptotic growth of $m(k)$" (even if the exponential lower bound is enough for you purpose, I think it is quite interesting in general to hold here). – Fedor Petrov May 27 '19 at 12:45

Ernie Croot showed that for all large $$N$$, every positive integer below $$\sum_{n\le N} \frac 1n - \Big(\frac{9}{2}+o(1)\Big) \frac{(\log \log N)^2}{\log N}$$ can be represented as a sum of unit fractions with denominator below $$N$$. Clearly no integer larger than $$\sum_{n\le N} 1/n$$ can be so represented. This gives the desired asymptotic growth for $$m(k)$$: namely $$m(k) = \exp(k-\gamma+ o(1)).$$