This follows from the colouring result of Croot [Cr03]. Croot's result allows for $n_k \leq e^{C^k}$ for some constant $C>1$ (simply taking $n_k$ to be the lowest common multiple of some interval $[1,C^k]$). Sawhney has observed that there is also a doubly exponential lower bound, and hence this bound is essentially sharp.

Indeed, we must trivially have $\sum_{d|n_k}1/d \geq k$, or else there is a greedy colouring as a counterexample. Since $\prod_{p}(1+1/p^2)$ is finite we must have $\prod_{p|n_k}(1+1/p)\gg k$. To achieve the minimal $\prod_{p|n_k}p$ we take the product of primes up to $T$ where $\prod_{p\leq T}(1+1/p)\gg k$; by Mertens theorems this implies $T\geq C^{k}$ for some constant $C>1$, and hence $n_k\geq \prod_{p\mid n_k}p\geq \exp(cC^k)$ for some $c>0$.

The answer is yes, as proved by Croot [Cr03].

See also [298].

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. This was improved by Liu and Sawhney [LiSa24] to \[\sum_{n\in A}\frac{1}{n}\gg (\log N)^{4/5+o(1)}.\] Erdős speculated 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.)

See also [46] and [298].

The current best bounds known are \[2^{c^{\frac{k}{\log k}}}\leq F(k) \leq c_0^{(\frac{1}{5}+o(1))2^k},\] where $c>0$ is some absolute constant and $c_0=1.26408\cdots$ is the 'Vardi constant'. The lower bound is due to Konyagin [Ko14] and the upper bound to Elsholtz and Planitzer [ElPl21].

Erdős and Graham write it is 'not difficult' to construct irrational $x$ such that this fails (although give no proof or reference, and it seems to still be an open problem to actually construct some such irrational $x$). Curtiss [Cu22] showed that this is true for $x=1$ and Erdős [Er50b] showed it is true for all $x=1/m$ with $m\geq 1$. Nathanson [Na23] has shown it is true for $x=a/b$ when $a\mid b+1$ and Chu [Ch23b] has shown it is true for a larger class of rationals; it is still unknown whether this is true for all rational $x>0$.

Without the 'eventually' condition this can fail for some rational $x$ (although Erdős [Er50b] showed it holds without the eventually for rationals of the form $1/m$). For example \[R_1(\tfrac{11}{24})=\frac{1}{3}\] but \[R_2(\tfrac{11}{24})=\frac{1}{4}+\frac{1}{5}.\]

Kovač [Ko24b] has proved that this is false - in fact as false as possible: the set of $x\in (0,\infty)$ for which the best underapproximations are eventually 'greedy' has Lebesgue measure zero. (It remains an open problem to give any explicit example of a number which is not eventually greedy, despite the fact that almost all numbers have this property.)

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 \[\frac{a}{n} = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}.\]

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.

Graham [Gr63] has proved this when $p(x)=x$. Cassels [Ca60] has proved that these conditions on the polynomial imply every sufficiently large integer is the sum of $p(n_i)$ with distinct $n_i$. Burr has proved this if $p(x)=x^k$ with $k\geq 1$ and if we allow $n_i=n_j$.

Alekseyev [Al19] has proved this when $p(x)=x^2$, for all $m>8542$. For example, \[1=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}\] and \[200 = 2^2+4^2+6^2+12^2.\]

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 $1=\sum \frac{1}{n_i}$.

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].

The answer is yes, proved by Croot [Cr01].

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.

For example, \[\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{20}=1.\] This is still open even if $\lvert I_2\rvert=1$. It is perhaps true with two intervals replaced by any $k$ intervals.

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

The smallest $b$ for each $a$ are listed at A375081 at the OEIS.

This was resolved in the affirmative by van Doorn [vD24], who proved $b=b(a)$ always exists, and in fact $b(a) \ll a$. Indeed, if $a\in (3^k,3^{k+1}]$ then one can take $b=2\cdot 3^{k+1}-1$. van Doorn also proves that $b(a)>a+(1/2-o(1))\log a$, and considers various generalisations of the original problem.

It seems likely that $b(a)\leq (1+o(1))a$, and perhaps even $b(a)\leq a+(\log a)^{O(1)}$.

Steinerberger has observed that the answer to the second question is trivially yes: for example, any $n$ which begins with a $2$ in base $3$ has $3\mid (a_n,L_n)$.

More generally, if the leading digit of $n$ in base $p$ is $p-1$ then $p\mid (a_n,L_n)$. There is in fact a necessary and sufficient condition: a prime $p\leq n$ divides $(a_n,L_n)$ if and only if $p$ divides the numerator of $1+\cdots+\frac{1}{k}$, where $k$ is the leading digit of $n$ in base $p$. This can be seen by writing \[a_n = \frac{L_n}{1}+\cdots+\frac{L_n}{n}\] and observing that the right-hand side is congruent to $1+\cdots+1/k$ modulo $p$. (The previous claim about $p-1$ follows immediately from Wolstenholme's theorem.)

This leads to a heuristic prediction (see for example a preprint of Shiu [Sh16]) of $\asymp\frac{x}{\log x}$ for the number of $n\in [1,x]$ such that $(a_n,L_n)=1$. In particular, there should be infinitely many $n$, but the set of such $n$ should have density zero. Unfortunately this heuristic is difficult to turn into a proof.

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

The answer is yes, as proved by Martin [Ma00], who in fact proved that if $B=\mathbb{N}\backslash A$ then, for all large $x$, \[\frac{\lvert B\cap [1,x]\rvert}{x}\asymp \frac{\log\log x}{\log x},\] and also gave an essentially complete description of $B$ as those integers which are small multiples of prime powers.

van Doorn has observed that if $n\in A$ (with $n>1$) then $2n\in A$ also, since if $\sum \frac{1}{m_i}=1$ then $\frac{1}{2}+\sum\frac{1}{2m_i}=1$ also.

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'.

Erdős and Graham [ErGr80] could show \[t(N)\ll\frac{N}{\log N},\] but had no idea of the true value of $t(N)$.

Solved by Liu and Sawhney [LiSa24] (up to $(\log\log N)^{O(1)}$), who proved that \[\frac{N}{(\log N)(\log\log N)^3(\log\log\log N)^{O(1)}}\ll t(N) \ll \frac{N}{\log N}.\]

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

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 [Bl21] (coupled with the trivial greedy approach) implies that $k(N)=(1-o(1))\log N$.

It was not even known for a long time whether this is $2^{cN}$ for some $c<1$ or $2^{(1+o(1))N}$. In fact the former is true, and the correct value of $c$ is now known.

• Steinerberger [St24] proved the relevant count is at most $2^{0.93N}$;
• Independently, Liu and Sawhney [LiSa24] gave both upper and lower bounds, proving that the count is \[2^{(0.91\cdots+o(1))N},\] where $0.91\cdots$ is an explicit number defined as the solution to a certain integral equation;
• Again independently this same asymptotic was proved (with a different proof) by Conlon, Fox, He, Mubayi, Pham, Suk, and Verstraëte [CFHMPSV24], who prove more generally, for any $x\in \mathbb{Q}_{>0}$, a similar expression for the number of $A\subseteq \{1,\ldots,N\}$ such that $\sum_{n\in A}\frac{1}{n}=x$;
• The above papers all appeared within weeks of each other in 2024; in 2017 a similar question (with $\leq 1$ rather than $=1$) was asked on MathOverflow by Mikhail Tikhomirov and proofs of the correct asymptotic were sketched by Lucia, RaphaelB4, and js21.

The answer is yes, proved by Bloom [Bl21].

See also [46] and [47].

There does not exist such a sequence, which follows from the positive solution to [298] by Bloom [Bl21].

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 $A(N)\geq (1-\frac{1}{e}+o(1))N$.

Liu and Sawhney [LiSa24] have proved that $A(N)=(1-1/e+o(1))N$.

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

Wouter van Doorn has given an elementary argument that proves \[f(N)\leq (25/28+o(1))N.\] Indeed, consider the sets $S_a=\{2a,3a,4a,6a,12a\}\cap [1,N]$ as $a$ ranges over all integers of the form $8^b9^cd$ with $(d,6)=1$. All such $S_a$ are disjoint and, if $A$ has no solutions to the given equation, then $A$ must omit at least two elements of $S_a$ when $a\leq N/12$ and at least one element of $S_a$ when $N/12<a\leq N/6$, and an elementary calculation concludes the proof.

Stijn Cambie and Wouter van Doorn have noted that, if we allow solutions to this equation with non-distinct $b_i$, then the size of the maximal set is at most $N/2$. Indeed, this is the classical threshold for the existence of some distinct $a,b\in A$ such that $a\mid b$.

See also [302] and [327].

The colouring version of this is [303], which was solved by Brown and Rödl [BrRo91]. One can take either $A$ to be all odd integers in $[1,N]$ or all integers in $[N/2,N]$ to show $f(N)\geq (1/2+o(1))N$.

Wouter van Doorn has proved, in an unpublished note, that \[f(N) \leq (9/10+o(1))N.\] Stijn Cambie has observed that \[f(N)\geq (5/8+o(1))N,\] taking $A$ to be all odd integers $\leq N/4$ and all integers in $[N/2,N]$.

Stijn Cambie has also observed that, if we allow $b=c$, then there is a solution to this equation when $\lvert A\rvert \geq (\tfrac{2}{3}+o(1))N$, since then there must exist some $n,2n\in A$.

See also [301] and [327].

The density version of this is [302]. This colouring version is true, as proved by Brown and Rödl [BrRo91].

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].

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

This was solved by Yokota [Yo88], who proved that \[D(b)\ll b(\log b)(\log\log b)^4(\log\log\log b)^2.\] This was improved by Liu and Sawhney [LiSa24] to \[D(b)\ll b(\log b)(\log\log b)^3(\log\log\log b)^{O(1)}.\]

For $n_i$ the product of three distinct primes, this is true when $b=1$, as proved by Butler, Erdős and Graham [BEG15] (this paper is perhaps Erdős' last paper, appearing 19 years after his death).

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 $Q$?

This was essentially solved by Croot [Cr99], who proved that if $f(N)$ is the smallest integer not representable then \[\left\lfloor\sum_{n\leq N}\frac{1}{n}-\frac{9}{2}(1+o(1))\frac{(\log\log N)^2}{\log N}\right\rfloor\leq f(N)\] and \[f(N)\leq \left\lfloor\sum_{n\leq N}\frac{1}{n}-\frac{1}{2}(1+o(1))\frac{(\log\log N)^2}{\log N}\right\rfloor.\] It follows that, if $m_N=\lfloor \sum_{n\leq N}\frac{1}{n}\rfloor$, then the set of integers representable is, for all $N$ sufficiently large, either $\{1,\ldots,m_N-1\}$ or $\{1,\ldots,m_N\}$.

The answer to the second question is no: there are at least $(1-o(1))\log N$ many such integers, which follows from a more precise result of Croot [Cr99], who showed that every integer at most \[\leq \sum_{n\leq N}\frac{1}{n}-(\tfrac{9}{2}+o(1))\frac{(\log\log N)^2}{\log N}\] can be so represented.

Liu and Sawhney [LiSa24] observed that the main result of Bloom [Bl21] implies a positive solution to this conjecture. They prove a more precise version, that if $(\log N)^{-1/7+o(1)}\leq \alpha \leq 1/2$ then there is some $S\subseteq A$ such that \[\frac{a}{b}=\sum_{n\in S}\frac{1}{n}\] with $a\leq b \leq \exp(O(1/\alpha))$. They also observe that the dependence $b\leq \exp(O(1/\alpha))$ is sharp.

It is trivially at least $1/[1,\ldots,N]$.

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

For example, \[\frac{1}{2}+\frac{1}{3}=1-\frac{1}{6}\] and \[\frac{1}{2}+\frac{1}{3}+\frac{1}{7}=1-\frac{1}{42}.\] It is clear that we must have $m=p_1\cdots p_k$, and hence in particular (up to ordering) there is at most one solution for each $m$. The integers $m$ for which there is such a solution are known as primary pseudoperfect numbers, and there are $8$ known, listed in A054377 at the OEIS.

This is true, and shown by Lim and Steinerberger [LiSt24] who proved that there exist infinitely many $n$ such that \[\epsilon(n)n^2\ll \left(\frac{\log\log n}{\log n}\right)^{1/2}.\] Erdős and Graham (and also Lim and Steinerberger) believe that the exponent of $2$ is best possible here, in that $\liminf \epsilon(n) n^{2+\delta}=\infty$ for all $\delta>0$.

$c_0$ is called the Vardi constant and the sequence $u_n$ is Sylvester's sequence.

In [ErGr80] this problem is stated with the sequence $u_1=1$ and $u_{n+1}=u_n(u_n+1)$, but Quanyu Tang has pointed out this is probably an error (since with that choice we do not have $\sum \frac{1}{u_n}=1$). This question with Sylvester's sequence is the most natural interpretation of what they meant to ask.

This is true, and was proved independently by Kamio [Ka25] and Li and Tang [LiTa25].

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.

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

Erdős and Straus [ErSt75] proved this when $A=\mathbb{N}$. Sattler [Sa75] proved this when $A$ is the set of odd numbers. For the squares $1$ must be excluded or the result is trivially false, since \[\sum_{k\geq 2}\frac{1}{k^2}<1.\]

Adenwalla has observed that a lower bound of \[\lvert A\rvert\geq (1-\tfrac{1}{e}+o(1))N\] follows from the main result of Croot [Cr01], which states that there exists some set of integers $B\subset [(\frac{1}{e}-o(1))N,N]$ such that $\sum_{b\in B}\frac{1}{b}=1$. Since the sum of $\frac{1}{m}$ for $m\in [c_1N,c_2N]$ is asymptotic to $\log(c_2/c_1)$ we must have $\lvert B\rvert \geq (1-\tfrac{1}{e}+o(1))N$.

We may then let $A=B\cup\{1\}$ and choose $\delta(n)=-1$ for all $n\in B$ and $\delta(1)=1$.

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].

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].

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

Wouter van Doorn has given an elementary argument that proves that if $A\subseteq \{1,\ldots,N\}$ has $\lvert A\rvert \geq (25/28+o(1))N$ then $A$ must contain $a\neq b$ with $a+b\mid ab$ (see the discussion in [301]).

See also [302].

Bleicher and Erdős conjecture the answer is no.