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 $m(k)$?
$\begingroup$
$\endgroup$
11
asked May 27, 2019 at 9:47
-
1$\begingroup$ I think this is somewhat related to harmonic numbers and the euler constant $\endgroup$– vidyarthiCommented May 27, 2019 at 9:58
-
1$\begingroup$ 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 $m(k)$. $\endgroup$– Fedor PetrovCommented May 27, 2019 at 10:16
-
3$\begingroup$ 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. $\endgroup$– Gerald EdgarCommented May 27, 2019 at 10:51
-
1$\begingroup$ ... but a comment there does mention the exponential lower bound. $\endgroup$– Gerald EdgarCommented May 27, 2019 at 11:47
-
3$\begingroup$ @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). $\endgroup$– Fedor PetrovCommented May 27, 2019 at 12:45
|
Show 6 more comments
1 Answer
$\begingroup$
$\endgroup$
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 $$ m(k) = \exp(k-\gamma+ o(1)). $$
