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SOLVED
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]. 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 $c>0$.

Does every finite colouring of the integers have a monochromatic solution to $1=\sum \frac{1}{n_i}$ with $2\leq n_1<\cdots <n_k$?
SOLVED - $100 If$\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. 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]. OPEN Let$F(k)$be the number of solutions to $1= \frac{1}{n_1}+\cdots+\frac{1}{n_k},$ where$1\leq n_1<\cdots<n_k$are distinct integers. Find good estimates for$F(k)$. 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]. SOLVED Let$x>0$be a real number. For any$n\geq 1$let $R_n(x) = \sum_{i=1}^n\frac{1}{m_i}<x$ be the maximal sum of$n$distinct unit fractions which is$<x$. Is it true that, for almost all$x$, for sufficiently large$n$, we have $R_{n+1}(x)=R_n(x)+\frac{1}{m},$ where$m$is minimal such that$m$does not appear in$R_n(x)$and the right-hand side is$<x$? (That is, are the best underapproximations eventually always constructed in a 'greedy' fashion?) 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.) Additional thanks to: Vjekoslav Kovac OPEN 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 $\frac{a}{n} = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$ OPEN 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. Additional thanks to: Zachary Hunter OPEN Let$p:\mathbb{Z}\to \mathbb{Z}$be a polynomial whose leading coefficient is positive and such that there exists no$d\geq 2$with$d\mid p(n)$for all$n\geq 1$. Is it true that, for all sufficiently large$m$, there exist integers$1\leq n_1<\cdots <n_k$such that $1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}$ and $m=p(n_1)+\cdots+p(n_k)?$ 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$n_i=n_j$. SOLVED 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$1=\sum \frac{1}{n_i}$. Additional thanks to: Zachary Hunter SOLVED 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]. Additional thanks to: Zachary Hunter SOLVED Let$k\geq 2$. Is it true that there exists an interval$I$of width$(e-1+o(1))k$and integers$n_1<\cdots<n_k\in I$such that $1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}?$ The answer is yes, proved by Croot [Cr01]. OPEN 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. OPEN Is it true that there are only finitely many pairs of intervals$I_1,I_2$such that $\sum_{n_1\in I_1}\frac{1}{n_1}+\sum_{n_2\in I_2}\frac{1}{n_2}\in \mathbb{N}?$ 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. Additional thanks to: Bhavik Mehta OPEN Is it true that, for all sufficiently large$k$, there exist finite intervals$I_1,\ldots,I_k\subset \mathbb{N}$with$\lvert I_i\rvert \geq 2$for$1\leq i\leq k$such that $1=\sum_{i=1}^k \sum_{n\in I_i}\frac{1}{n}?$ OPEN 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}.$ The smallest$b$for each$a$are listed at A375081 at the OEIS. Additional thanks to: Ralf Stephan OPEN Let$n\geq 1$and define$L_n$to be the least common multiple of$\{1,\ldots,n\}$and$a_n$by $\sum_{1\leq k\leq n}\frac{1}{k}=\frac{a_n}{L_n}.$ Is it true that$(a_n,L_n)=1$and$(a_n,L_n)>1$both occur for infinitely many$n$? 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. Additional thanks to: Stefan Steinerberger OPEN 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. Additional thanks to: Zachary Hunter OPEN 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'. Additional thanks to: Zachary Hunter SOLVED 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] 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}.$ OPEN 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}.$ SOLVED 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$. Additional thanks to: Zachary Hunter and Mehtaab Sawhney SOLVED Let$N\geq 1$. How many$A\subseteq \{1,\ldots,N\}$are there such that$\sum_{n\in A}\frac{1}{n}=1$? 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. Additional thanks to: Zachary Chase SOLVED Does every set$A\subseteq \mathbb{N}$of positive density contain some finite$S\subset A$such that$\sum_{n\in S}\frac{1}{n}=1$? The answer is yes, proved by Bloom [Bl21]. See also [46] and [47]. SOLVED Is there an infinite sequence$a_1<a_2<\cdots $such that$a_{i+1}-a_i=O(1)$and no finite sum of$\frac{1}{a_i}$is equal to$1$? There does not exist such a sequence, which follows from the positive solution to [298] by Bloom [Bl21]. SOLVED 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$. Liu and Sawhney [LiSa24] have proved that$A(N)=(1-1/e+o(1))N$. Additional thanks to: Zachary Hunter OPEN 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$. Additional thanks to: Zachary Hunter OPEN 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. Additional thanks to: Zachary Hunter and Mehtaab Sawhney SOLVED Is it true that in any finite colouring of the integers there exists a monochromatic solution to $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$ with distinct$a,b,c$? The density version of this is [302]. This colouring version is true, as proved by Brown and Rödl [BrRo91]. OPEN 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]. Additional thanks to: Zachary Hunter SOLVED 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.$ 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)}.$ OPEN 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 [BEG15] (this paper is perhaps Erdős' last paper, appearing 19 years after his death). OPEN 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$? OPEN Let$N\geq 1$. What is the smallest integer not representable as the sum of distinct unit fractions with denominators from$\{1,\ldots,N\}$? Is it true that the set of integers not representable as such has the shape$[m,\infty)$for some$m$? SOLVED 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? 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. Additional thanks to: Zachary Hunter and Mehtaab Sawhney SOLVED Let$\alpha >0$and$N\geq 1$. Is it true that for any$A\subseteq \{1,\ldots,N\}$with$\lvert A\rvert \geq \alpha N$there exists some$S\subseteq A$such that $\frac{a}{b}=\sum_{n\in S}\frac{1}{n}$ with$a\leq b =O_\alpha(1)$? 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. OPEN What is the minimal value of$\lvert 1-\sum_{n\in A}\frac{1}{n}\rvert$as$A$ranges over all subsets of$\{1,\ldots,N\}$which contain no$S$such that$\sum_{n\in S}\frac{1}{n}=1$? Is it $e^{-(c+o(1))N}$ for some constant$c\in (0,1)$? It is trivially at least$1/[1,\ldots,N]$. OPEN 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$. Additional thanks to: Mehtaab Sawhney OPEN Are there infinitely many solutions to $\frac{1}{p_1}+\cdots+\frac{1}{p_k}=1-\frac{1}{m},$ where$m\geq 2$is an integer and$p_1<\cdots<p_k$are distinct primes? For example, $\frac{1}{2}+\frac{1}{3}=1-\frac{1}{6}.$ SOLVED Let$n\geq 1$and let$m$be minimal such that$\sum_{n\leq k\leq m}\frac{1}{k}\geq 1$. We define $\epsilon(n) = \sum_{n\leq k\leq m}\frac{1}{k}-1.$ How small can$\epsilon(n)$be? Is it true that $\liminf n^2\epsilon(n)=0?$ 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$. OPEN Let$u_1=2$and$u_{n+1}=u_n^2-u_n+1$. Let$a_1<a_2<\cdots $be any other sequence with$\sum \frac{1}{a_k}=1$. Is it true that $\liminf a_n^{1/2^n}<\lim u_n^{1/2^n}=c_0=1.264085\cdots?$$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$). This question with Sylvester's sequence is the most natural interpretation of what they meant to ask. Additional thanks to: Quanyu Tang SOLVED Is it true that if$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. Additional thanks to: Tom Stobart OPEN 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 $\frac{1}{2}-\frac{1}{3}-\frac{1}{4}=-\frac{1}{12}.$ Additional thanks to: Zachary Chase OPEN Let$A\subseteq \mathbb{N}$be an infinite arithmetic progression and$f:A\to \{-1,1\}$be a non-constant function. Must there exist a finite$S\subset A$such that $\sum_{n\in S}\frac{f(n)}{n}=0?$ What about if$A$is an arbitrary set of positive density? What if$A$is the set of squares excluding$1$? Erdős and Straus [ErSt75] proved this when$A=\mathbb{N}$. Sattler [Sat75] proved this when$A$is the set of odd numbers. For the squares$1$must be excluded or the result is trivially false, since $\sum_{k\geq 2}\frac{1}{k^2}<1.$ OPEN What is the size of the largest$A\subseteq \{1,\ldots,N\}$such that there is a function$\delta:A\to \{-1,1\}$such that $\sum_{n\in A}\frac{\delta_n}{n}=0$ and $\sum_{n\in A'}\frac{\delta_n}{n}\neq 0$ for all$A'\subsetneq A$? OPEN 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]. Additional thanks to: Boris Alexeev and Dustin Mixon OPEN 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]. Additional thanks to: Boris Alexeev, Zachary Hunter, and Dustin Mixon OPEN Suppose$A\subseteq \{1,\ldots,N\}$is such that if$a,b\in A$and$a\neq b$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$a+b\mid ab$. SOLVED 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. Steinerberger has pointed out that as written this problem is trivial: simply take some lacunary$A$whose prime factors are restricted (e.g.$A=\{1,2,4,8,\ldots,\}$) - clearly any finite sum of the shape$\sum_{a\in A'}\frac{1}{a}$can only form a rational with denominator divisible by one of these restricted set of primes. This is puzzling, since Erdős and Graham were very aware of this kind of obstruction, so it's a strange thing to miss. I assume that there was some unwritten extra assumption intended (e.g.$A\$ contains a multiple of every integer).