12 solved out of 48 shown

Let $k\geq 2$. Is there an integer $n_k$ such that, if $D=\{ 1<d<n_k : d\mid n_k\}$, then for any $k$-colouring of $D$ there is a monochromatic subset $D'\subseteq D$ such that $\sum_{d\in D'}\frac{1}{d}=1$?

This follows from the colouring result of Croot [Cr03]. Obtaining better quantitative aspects seems interesting still -- note that Croot's result allows for $n_k \leq e^{C^k}$ for some constant $C>1$. Is there a better bound?

$100

If $\delta>0$ and $N$ is sufficiently large in terms of $\delta$, and $A\subseteq\{1,\ldots,N\}$ is such that $\sum_{a\in A}\frac{1}{a}>\delta \log N$ then must there exist $S\subseteq A$ such that $\sum_{n\in S}\frac{1}{n}=1$?

Solved by Bloom [Bl21], who showed that the quantitative threshold
\[\sum_{n\in A}\frac{1}{n}\gg \frac{\log\log\log N}{\log\log N}\log N\]
is sufficient. Erdős further speculates that perhaps even $\gg (\log\log N)^2$ might be sufficient. (A construction of Pomerance, as discussed in the appendix of [Bl21], shows that this would be best possible.)

For every $n\geq 2$ there exist distinct integers $1\leq x<y<z$ such that
\[\frac{4}{n} = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}.\]

The Erdős-Straus conjecture. The existence of a representation of $4/n$ as the sum of at most four distinct unit fractions follows trivially from a greedy algorithm.

Schinzel conjectured the generalisation that, for any fixed $a$, if $n$ is sufficiently large in terms of $a$ then there exist distinct integers $1\leq x<y<z$ such that \[\frac{a}{n} = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}.\]

Let $A\subseteq \mathbb{N}$ be an infinite set and consider the following greedy algorithm for a rational $x\in (0,1)$: choose the minimal $n\in A$ such that $n\geq 1/x$ and repeat with $x$ replaced by $x-\frac{1}{n}$. If this terminates after finitely many steps then this produces a representation of $x$ as the sum of distinct unit fractions with denominators from $A$.

Does this process always terminate if $x$ has odd denominator and $A$ is the set of odd numbers? More generally, for which pairs $x$ and $A$ does this process terminate?

In 1202 Fibonacci observed that this process terminates for any $x$ when $A=\mathbb{N}$. The problem when $A$ is the set of odd numbers is due to Stein.

Graham [Gr64b] has shown that $\frac{m}{n}$ is the sum of distinct unit fractions with denominators $\equiv a\pmod{d}$ if and only if \[\left(\frac{n}{(n,(a,d))},\frac{d}{(a,d)}\right)=1.\] Does the greedy algorithm always terminate in such cases?

Graham [Gr64c] has also shown that $x$ is the sum of distinct unit fractions with square denominators if and only if $x\in [0,\pi^2/6-1)\cup [1,\pi^2/6)$. Does the greedy algorithm for this always terminate? Erdős and Graham believe not - indeed, perhaps it fails to terminate almost always.

Let $f(k)$ be the maximal value of $n_1$ such that there exist $n_1<n_2<\cdots <n_k$ with
\[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}.\]
Is it true that
\[f(k)=(1+o(1))\frac{k}{e-1}?\]

The upper bound $f(k) \leq (1+o(1))\frac{k}{e-1}$ is trivial since for any $u\geq 1$ we have
\[\sum_{u\leq n\leq eu}\frac{1}{n}=1+o(1),\]
and hence if $f(k)=u$ then we must have $k\geq (e-1-o(1))u$.
Essentially solved by Croot [Cr01], who showed that for any $N>1$ there exists some $k\geq 1$ and
\[N<n_1<\cdots <n_k \leq (e+o(1))N\]
with $1=\sum \frac{1}{n_i}$.

Let $f(k)$ be the minimal value of $n_k$ such that there exist $n_1<n_2<\cdots <n_k$ with
\[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}.\]
Is it true that
\[f(k)=(1+o(1))\frac{e}{e-1}k?\]

It is trivial that $f(k)\geq (1+o(1))\frac{e}{e-1}k$, since for any $u\geq 1$
\[\sum_{e\leq n\leq eu}\frac{1}{n}= 1+o(1),\]
and so if $eu\approx f(k)$ then $k\leq \frac{e-1}{e}f(k)$. Proved by Martin [Ma00].

Let $k\geq 2$. Is it true that, for any distinct integers $n_1<\cdots <n_k$ such that
\[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}\]
we must have $\max(n_{i+1}-n_i)\geq 3$?

The example $1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ shows that $3$ would be best possible here. The lower bound of $\geq 2$ is equivalent to saying that $1$ is not the sum of reciprocals of consecutive integers, proved by Erdős [Er32].

This conjecture would follow for all but at most finitely many exceptions if it were known that, for all large $N$, there exists a prime $p\in [N,2N]$ such that $\frac{p+1}{2}$ is also prime.

Let $a\geq 1$. Must there exist some $b>a$ such that
\[\sum_{a\leq n\leq b}\frac{1}{n}=\frac{r_1}{s_1}\textrm{ and }\sum_{a\leq n\leq b+1}\frac{1}{n}=\frac{r_2}{s_2},\]
with $(r_i,s_i)=1$ and $s_2<s_1$? If so, how does this $b(a)$ grow with $a$?

For example,
\[\sum_{3\leq n\leq 5}\frac{1}{n} = \frac{47}{60}\textrm{ and }\sum_{3\leq n\leq 6}\frac{1}{n}=\frac{19}{20}.\]

Let $A$ be the set of $n\in \mathbb{N}$ such that there exist $1\leq m_1<\cdots <m_k=n$ with $\sum\tfrac{1}{m_i}=1$. Explore $A$. In particular,

- Does $A$ have density $1$?
- What are those $n\in A$ not divisible by any $d\in A$ with $1<d<n$?

Straus observed that $A$ is closed under multiplication. Furthermore, it is easy to see that $A$ does not contain any prime power.

Let $k\geq 1$ and let $v(k)$ be the minimal integer which does not appear as some $n_i$ in a solution to
\[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}\]
with $1\leq n_1<\cdots <n_k$. Estimate the growth of $v(k)$.

Results of Bleicher and Erdős [BlEr75] imply $v(k) \gg k!$. It may be that $v(k)$ grows doubly exponentially in $\sqrt{k}$ or even $k$.

An elementary inductive argument shows that $n_k\leq ku_k$ where $u_1=1$ and $u_{i+1}=u_i(u_i+1)$, and hence \[v(k) \leq kc_0^{2^k},\] where \[c_0=\lim_n u_n^{1/2^n}=1.26408\cdots\] is the 'Vardi constant'.

Let $N\geq 1$ and let $t(N)$ be the least integer $t$ such that there is no solution to
\[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}\]
with $t=n_1<\cdots <n_k\leq N$. Estimate $t(N)$.

Erdős and Graham [ErGr80] claim a bound of
\[t(N)\ll\frac{N}{\log N},\]
but have no idea of the true value of $t(N)$.

Let $N\geq 1$ and let $k(N)$ denote the smallest $k$ such that there exist $N\leq n_1<\cdots <n_k$ with
\[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}.\]
Is it true that
\[\lim_{N\to \infty} k(N)-(e-1)N=\infty?\]

Erdős and Straus [ErSt71b] have proved the existence of some constant $c>0$ such that
\[-c < k(N)-(e-1)N \ll \frac{N}{\log N}.\]

Let $N\geq 1$ and let $k(N)$ be maximal such that there are $k$ disjoint $A_1,\ldots,A_k\subseteq \{1,\ldots,N\}$ with $\sum_{n\in A_i}\frac{1}{n}=1$ for all $i$. Estimate $k(N)$. Is it true that $k(N)=o(\log N)$?

More generally, how many disjoint $A_i$ can be found in $\{1,\ldots,N\}$ such that the sums $\sum_{n\in A_i}\frac{1}{n}$ are all equal? Using sunflowers it can be shown that there are at least $N\exp(-O(\sqrt{\log N}))$ such sets.

Hunter and Sawhney have observed that Theorem 3 of Bloom [Bl23] (coupled with the trivial greedy approach) implies that $k(N)=(1-o(1))\log N$.

Let $A(N)$ denote the maximal cardinality of $A\subseteq \{1,\ldots,N\}$ such that $\sum_{n\in S}\frac{1}{n}\neq 1$ for all $S\subseteq A$. Estimate $A(N)$.

Erdős and Graham believe the answer is $A(N)=(1+o(1))N$. Croot [Cr03] disproved this, showing the existence of some constant $c<1$ such that $A(N)<cN$ for all large $N$. It is trivial that $A(N)\geq (1-\frac{1}{e}+o(1))N$.

What is the size of the largest $A\subseteq \{1,\ldots,N\}$ such that for any $a,b_1,\ldots,b_k\in A$ with $k\geq 2$ we have
\[\frac{1}{a}\neq \frac{1}{b_1}+\cdots+\frac{1}{b_k}?\]
Is there a constant $c>1/2$ such that $\lvert A\rvert >cN$ is possible for all large $N$?

The example $A=(N/2,N]\cap \mathbb{N}$ shows that $\lvert A\rvert\geq N/2$.

Is it true that, for any $\delta>1/2$, if $N$ is sufficiently large and $A\subseteq \{1,\ldots,N\}$ has $\lvert A\rvert \geq \delta N$ then there exist $a,b,c\in A$ such that
\[\frac{1}{a}=\frac{1}{b}+\frac{1}{c}.\]

The colouring version of this is [303], which was solved by Brown and Rödl [BrRo91].

The possible alternative question, that if $A\subseteq \mathbb{N}$ is a set of positive lower density then must there exist $a,b,c\in A$ such that \[\frac{1}{a}=\frac{1}{b}+\frac{1}{c},\] has a negative answer, taking for example $A$ to be the union of $[5^k,(1+\epsilon)5^k]$ for large $k$ and sufficiently small $\epsilon>0$. This was observed by Hunter and Sawhney.

For integers $1\leq a<b$ let $N(a,b)$ denote the minimal $k$ such that there exist integers $1<n_1<\cdots<n_k$ with
\[\frac{a}{b}=\frac{1}{n_1}+\cdots+\frac{1}{n_k}.\]
Estimate $N(b)=\max_{1\leq a<b}N(a,b)$. Is it true that $N(b) \ll \log\log b$?

Erdős [Er50c] proved that
\[\log\log b \ll N(b) \ll \frac{\log b}{\log\log b}.\]
The upper bound was improved by Vose [Vo85] to
\[N(b) \ll \sqrt{\log b}.\]
One can also investigate the average of $N(a,b)$ for fixed $b$, and it is known that
\[\frac{1}{b}\sum_{1\leq a<b}N(a,b) \gg \log\log b.\]

Related to [18].

For integers $1\leq a<b$ let $D(a,b)$ be the minimal value of $n_k$ such that there exist integers $1\leq n_1<\cdots <n_k$ with
\[\frac{a}{b}=\frac{1}{n_1}+\cdots+\frac{1}{n_k}.\]
Estimate $D(b)=\max_{1\leq a<b}D(a,b)$. Is it true that
\[D(b) \ll b(\log b)^{1+o(1)}?\]

Bleicher and Erdős [BlEr76] have shown that
\[D(b)\ll b(\log b)^2.\]
If $b=p$ is a prime then
\[D(p) \gg p\log p.\]

Let $a/b\in \mathbb{Q}_{>0}$ with $b$ squarefree. Are there integers $1<n_1<\cdots<n_k$, each the product of two distinct primes, such that
\[\frac{a}{b}=\frac{1}{n_1}+\cdots+\frac{1}{n_k}?\]

For $n_i$ the product of three distinct primes, this is true when $b=1$, as proved by Butler, Erdős and Graham [BuErGr15] (this paper is perhaps Erdős' last paper, appearing 19 years after his death).

Are there two finite sets of primes $P,Q$ such that
\[1=\left(\sum_{p\in P}\frac{1}{p}\right)\left(\sum_{q\in Q}\frac{1}{q}\right)?\]

Asked by Barbeau [Ba76]. Can this be done if we drop the requirement that all $p\in P$ are prime and just ask for them to be relatively coprime, and similarly for $Q$?

Let $N\geq 1$. How many integers can be written as the sum of distinct unit fractions with denominators from $\{1,\ldots,N\}$? Are there $o(\log N)$ such integers?

Hunter and Sawhney have observed that Theorem 3 of Bloom [Bl23] implies the answer to the second question is no: in fact all integers $m\leq (1-o(1))\log N$ can be written as the sum of distinct unit fractions with denominators from $\{1,\ldots,N\}$.

Does there exist some $c>0$ such that, for any $K>1$, whenever $A$ is a sufficiently large finite multiset of integers with $\sum_{n\in A}\frac{1}{n}>K$ there exists some $S\subseteq A$ such that
\[1-e^{-cK} < \sum_{n\in S}\frac{1}{n}\leq 1?\]

Erdős and Graham knew this with $e^{-cK}$ replaced by $c/K^2$.

Is it true that if $A\subset \mathbb{N}\backslash\{1\}$ is a finite set with $\sum_{n\in A}\frac{1}{n}<2$ then there is a partition $A=A_1\sqcup A_2$ such that
\[\sum_{n\in A_i}\frac{1}{n}<1\]
for $i=1,2$?

This is not true if $A$ is a multiset, for example $2,3,3,5,5,5,5$.

This is not true in general, as shown by Sándor [Sa97], who observed that the proper divisors of $120$ form a counterexample. More generally, Sándor shows that for any $n\geq 2$ there exists a finite set $A\subseteq \mathbb{N}\backslash\{1\}$ with $\sum_{k\in A}\frac{1}{k}<n$ and no partition into $n$ parts each of which has $\sum_{k\in A_i}\frac{1}{k}<1$.

The minimal counterexample is $\{2,3,4,5,6,7,10,11,13,14,15\}$, found by Tom Stobart.

Is there some constant $c>0$ such that for every $n\geq 1$ there exists some $\delta_k\in \{-1,0,1\}$ for $1\leq k\leq n$ with
\[0< \left\lvert \sum_{1\leq k\leq n}\frac{\delta_k}{k}\right\rvert < \frac{c}{2^n}?\]
Is it true that for sufficiently large $n$, for any $\delta_k\in \{-1,0,1\}$,
\[\left\lvert \sum_{1\leq k\leq n}\frac{\delta_k}{k}\right\rvert > \frac{1}{[1,\ldots,n]}\]
whenever the left-hand side is not zero?

Inequality is obvious for the second claim, the problem is strict inequality. This fails for small $n$, for example
\[\frac{1}{2}-\frac{1}{3}-\frac{1}{4}=-\frac{1}{12}.\]

Let $S(N)$ count the number of distinct sums of the form $\sum_{n\in A}\frac{1}{n}$ for $A\subseteq \{1,\ldots,N\}$. Estimate $S(N)$.

Bleicher and Erdős [BlEr75] proved the lower bound
\[\frac{N}{\log N}\prod_{i=3}^k\log_iN\leq \frac{\log S(N)}{\log 2},\]
valid for $k\geq 4$ and $\log_kN\geq k$, and also [BlEr76b] proved the upper bound
\[\log S(N)\leq \log_r N\left(\frac{N}{\log N} \prod_{i=3}^r \log_iN\right),\]
valid for $r\geq 1$ and $\log_{2r}N\geq 1$. (In these bounds $\log_in$ denotes the $i$-fold iterated logarithm.)

See also [321].

What is the size of the largest $A\subseteq \{1,\ldots,N\}$ such that all sums $\sum_{n\in S}\frac{1}{n}$ are distinct for $S\subseteq A$?

Let $R(N)$ be the maximal such size. Results of Bleicher and Erdős from [BlEr75] and [BlEr76b] imply that
\[\frac{N}{\log N}\prod_{i=3}^k\log_iN\leq R(N)\leq \frac{1}{\log 2}\log_r N\left(\frac{N}{\log N} \prod_{i=3}^r \log_iN\right),\]
valid for any $k\geq 4$ with $\log_kN\geq k$ and any $r\geq 1$ with $\log_{2r}N\geq 1$. (In these bounds $\log_in$ denotes the $i$-fold iterated logarithm.)

See also [320].

Suppose $A\subseteq \{1,\ldots,N\}$ is such that if $a,b\in A$ then $a+b\nmid ab$. Can $A$ be 'substantially more' than the odd numbers?

What if $a,b\in A$ with $a\neq b$ implies $a+b\nmid 2ab$? Must $\lvert A\rvert=o(N)$?

The connection to unit fractions comes from the observation that $\frac{1}{a}+\frac{1}{b}$ is a unit fraction if and only if $a+b\mid ab$.

Let $A\subseteq \mathbb{N}$ be a lacunary sequence (so that $A=\{a_1<a_2<\cdots\}$ and there exists some $\lambda>1$ such that $a_{n+1}/a_n\geq \lambda$ for all $n\geq 1$). Must
\[\left\{ \sum_{a\in A'}\frac{1}{a} : A'\subseteq A\textrm{ finite}\right\}\]
contain all rationals in some open interval?

Bleicher and Erdős conjecture the answer is no.