Erdős called this 'perhaps my first serious problem'. The powers of $2$ show that $2^n$ would be best possible here. The trivial lower bound is $N \gg 2^{n}/n$, since all $2^n$ distinct subset sums must lie in $\{1,\ldots,Nn\}$. Erdős and Moser [Er56] proved \[ N\geq (\tfrac{1}{4}-o(1))\frac{2^n}{\sqrt{n}}.\] A number of improvements of the constant have been given (see [St23] for a history), with the current record $\sqrt{2/\pi}$ due to Dubroff, Fox, and Xu [DFX21].

In [ErGr80] the generalisation where $A\subseteq (0,N]$ is a set of real numbers such that the subset sums all differ by at least $1$ is proposed, with the same conjectured bound.

Described by Erdős as 'perhaps my favourite problem'. Hough [Ho15] has shown the answer is no: the smallest modulus must be at most $10^{18}$.

This is essentially asking for good bounds on $r_k(N)$, the size of the largest subset of $\{1,\ldots,N\}$ without a non-trivial $k$-term arithmetic progression. For example, a bound like \[r_k(N) \ll_k \frac{N}{(\log N)(\log\log N)^2}\] would be sufficient.

Even the case $k=3$ is non-trivial, but was proved by Bloom and Sisask [BlSi20]. Much better bounds for $r_3(N)$ were subsequently proved by Kelley and Meka [KeMe23]. Green and Tao [GrTa17] have proved $r_4(N)\ll N/(\log N)^{c}$ for some small constant $c>0$. The best bound available for general $k$ is due to Gowers [Go01], \[r_k(N) \ll \frac{N}{(\log\log N)^{c_k}},\] where $c_k>0$ is a small constant depending on $k$.

Curiously, Erdős [Er83c] thought this conjecture was the 'only way to approach' the conjecture that there are arbitrarily long arithmetic progressions of prime numbers, now a theorem due to Green and Tao [GrTa08].

See also [142].

The peculiar quantitative form of Erdős' question was motivated by an old result of Rankin [Ra38], which showed the existence of some constant $C>0$ such that the claim holds. Solved by Maynard [Ma16] and Ford, Green, Konyagin, and Tao [FoGrKoTa16]. The best bound available, due to all five authors [FoGrKoMaTa18], is that there are infinitely many $n$ such that \[p_{n+1}-p_n\gg \frac{\log\log n\log\log\log\log n}{\log\log \log n}\log n.\] The likely truth is a lower bound like $\gg(\log n)^2$. In [Er97c] Erdős revised the value of this problem to \$5000 and reserved the \$10000 for a lower bound of $>(\log n)^{1+c}$ for some $c>0$.

Let $S$ be the set of limit points of $(p_{n+1}-p_n)/\log n$. This problem asks whether $S=[0,\infty]$. Although this conjecture remains unproven, a lot is known about $S$. Some highlights:

• $\infty\in S$ by Westzynthius' result [We31] on large prime gaps,
• $0\in S$ by the work of Goldston, Pintz, and Yildirim [GPY09] on small prime gaps,
• Erdős [Er55] and Ricci [Ri56] independently showed that $S$ has positive Lebesgue measure,
• Hildebrand and Maier [HiMa88] showed that $S$ contains arbitrarily large (finite) numbers,
• Pintz [Pi16] showed that there exists some small constant $c>0$ such that $[0,c]\subset S$,
• Banks, Freiberg, and Maynard [BFM16] showed that at least $12.5\%$ of $[0,\infty)$ belongs to $S$,
• Merikoski [Me20] showed that at least $1/3$ of $[0,\infty)$ belongs to $S$, and that $S$ has bounded gaps.

Conjectured by Erdős and Turán [ErTu48]. Shockingly Erdős offered \$25000 for a disproof of this, but as he comments, it 'is certainly true'.

Indeed, the answer is yes, as proved by Banks, Freiberg, and Turnage-Butterbaugh [BFT15] with an application of the Maynard-Tao machinery concerning bounded gaps between primes [Ma15]. They in fact prove that, for any $m\geq 1$, there are infinitely many $n$ such that \[d_n<d_{n+1}<\cdots <d_{n+m}\] and infinitely many $n$ such that \[d_n> d_{n+1}>\cdots >d_{n+m}.\]

Asked by Erdős and Selfridge (sometimes also with Schinzel). They also ask can there be such that no two of the moduli divide each other, or where all the moduli are odd and square-free?

Selfridge has shown (as reported in [Sc67]) that such a covering system exists if a covering system exists with moduli $n_1,\ldots,n_k$ such that no $n_i$ divides any other $n_j$.

Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST21] have proved that if the moduli are all squarefree then at least one must be even.

Conjectured by Erdős and Graham, who also ask about a density-type version: for example, is \[\sum_{\substack{a\in A\\ a>N}}\frac{1}{a}\gg \log N\] a sufficient condition for $A$ to contain the moduli of a covering system? The answer (to both colouring and density versions) is no, due to the result of Hough [Ho15] on the minimum size of a modulus in a covering system - in particular one could colour all integers $<10^{18}$ different colours and all other integers a new colour.

Crocker [Cr71] has proved there are are $\gg\log\log N$ such integers in $\{1,\ldots,N\}$. Erdős believes this cannot be proved by covering systems, i.e. integers of the form $p+2^k+2^l$ exist in every infinite arithmetic progression.

Erdős described this as 'probably unattackable'. In [ErGr80] Erdős and Graham suggest that no such $k$ exists. Gallagher [Ga75] has shown that for any $\epsilon>0$ there exists $k(\epsilon)$ such that the set of integers which are the sum of a prime and at most $k(\epsilon)$ many powers of 2 has lower density at least $1-\epsilon$.

Odlyzko has checked this up to $10^7$. Granville and Soundararajan [GrSo98] have proved that this is very related to the problem of finding primes $p$ for which $2^p\equiv 2\pmod{p^2}$ (for example this conjecture implies there are infinitely many such $p$).

Erdős thinks that proving this with two powers of 2 is perhaps easy.

Asked by Erdős and Sárközy, who proved that $A$ must have density $0$. The set of $p^2$, where $p\equiv 3\pmod{4}$ is prime, has this property. Note that such a set cannot contain a three-term arithmetic progression, and hence by the bound of Kelley and Meka [KeMe23] we have \[\lvert A\cap\{1,\ldots,N\}\rvert\ll \exp(-O((\log N)^{1/12}))N\] for all large $N$.

Asked by Erdős and Sárközy, who observed that $[2N/3,N]\cap \mathbb{N}$ is such a set. The answer is yes, as proved by Bedert [Be23].

Asked by Erdős, Sárközy, and Szemerédi, who constructed a set which shows this would be the best possible.

Erdős suggested that a computer could be used to explore this, and did not see any other method to attack this.

Tao [Ta23] has proved that this series does converge assuming a strong form of the Hardy-Littlewood prime tuples conjecture.

Erdős called this conjecture 'rather silly'. Using covering congruences Erdős [Er50] proved that the set of such odd integers contains an infinite arithmetic progression.

The first prime without this property is $97$.

It is easy to see that almost all numbers are not practical. This may be true with $m=n!$. Erdős originally showed that $h(n!) <n$. Vose [Vo85] has shown that $h(n!)\ll n^{1/2}$.

Related to [304].

Asked by Erdős and Tenenbaum. Ruzsa has found a counterexample. Tenenbaum asked the weaker variant (still open) where for every $\epsilon>0$ there is some $k=k(\epsilon)$ such that at least $1-\epsilon$ density of all integers have a divisor of the form $a+k$ for some $a\in A$.

By a simple averaging argument the set of moduli $[m_1,m_2]\cap \mathbb{N}$ has a choice of residue classes which form an $\epsilon(m_1,m_2)$-almost covering system with \[\epsilon(m_1,m_2)=\prod_{m_1\leq m\leq m_2}(1-1/m).\] A $0$-covering system is just a covering system, and so by Hough [Ho15] these only exist for $n_1<10^{18}$. [NOTE: This is my best attempt at recovering problem 5 from [Er95], which doesn't make sense as written.]

Conjectured by Erdős and Turán. They also suggest the stronger conjecture that $\limsup 1_A\ast 1_A(n)/\log n>0$. Another stronger conjecture would be that $\lvert A\cap [1,N]\rvert \gg N^{1/2}$ for all large $N$ implies $\limsup 1_A\ast 1_A(n)=\infty$. Erdős and Sárközy conjectured the stronger version that if $A=\{a_1<a_2<\cdots\}$ and $B=\{b_1<b_2<\cdots\}$ with $a_n/b_n\to 1$ such that $A+B=\mathbb{N}$ then $\limsup 1_A\ast 1_B(n)=\infty$.

The existence of such a set was asked by Sidon to Erdős in 1932. Erdős (eventually) proved the existence of such a set using probabilistic methods. This problem asks for a constructive solution.

It may even be true that $h(N)=N^{1/2}+O(1)$, but Erdős remarks this is perhaps too optimistic. Erdős and Turán [ErTu41] proved an upper bound of $N^{1/2}+O(N^{1/4})$, with an alternative proof by Lindström [Li69]. Balogh, Füredi, and Roy [BFR21] improved the bound in the error term to $0.998N^{1/4}$, which has been further optimised by O'Bryant [OB22] to yield \[h(N)\leq N^{1/2}+0.99703N^{1/4}\] for sufficiently large $N$.

Proved by Lorentz [Lo54].

Erdős [Er54] proved by the probabilistic method that such a set $A$ exists with $\lvert A\cap\{1,\ldots,N\}\rvert\ll (\log N)^2$.

Erdős observed that this value is finite and $>1$.

Erdős proved this is true with $k$ replaced by $2k$ in the denominator.

The minimum overlap problem. The example (with $N$ even) $A=\{N/2+1,\ldots,3N/2\}$ shows that $c\leq 1/2$ (indeed, Erdős initially conjectured that $c=1/2$). The lower bound of $c\geq 1/4$ is trivial, and Scherk improved this to $1-1/\sqrt{2}=0.29\cdots$. The current records are \[0.379005 < c < 0.380926\cdots,\] the lower bound due to White [Wh22] and the upper bound due to Haugland [Ha16].

The answer is no by Ruzsa [Ru87], who proved that if $A$ is an essential component then there exists some constant $c>0$ such that $\lvert A\cap \{1,\ldots,N\}\rvert \geq (\log N)^{1+c}$ for all large $N$.

The trivial greedy construction achieves $\gg N^{1/3}$. The current best bound of $\gg N^{\sqrt{2}-1+o(1)}$ is due to Ruzsa [Ru98]. Erdős proved that for every infinite Sidon set $A$ we have \[\liminf \frac{\lvert A\cap \{1,\ldots,N\}\rvert}{N^{1/2}}=0,\] and also that there is a set $A\subset \mathbb{N}$ with $\lvert A\cap \{1\ldots,N\}\rvert \gg_\epsilon N^{1/2-\epsilon}$ such that $1_A\ast 1_A(n)=O(1)$.

It is possible that this is true even with $g(N)=O(1)$, from which the Erdős-Turán conjecture would follow.

Erdős proved that if the pairwise sums $a+b$ are all distinct aside from the trivial coincidences then \[\liminf \frac{\lvert A\cap \{1,\ldots,N\}\rvert}{N^{1/2}}=0.\]

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?

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

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.)

This would follow immediately from the twin prime conjecture. The answer is yes, proved by Ford, Luca, and Pomerance [FLP10].

Erdős remarks that the last conjecture is probably easy, and that similar questions can be asked about $\sigma(n)$.

Solved by Tao [Ta23b], who proved that \[ \lvert A\rvert \leq \left(1+O\left(\frac{(\log\log x)^5}{\log x}\right)\right)\pi(x).\]

Carmichael has asked whether there is an integer $t$ for which $\phi(n)=t$ has exactly one solution. Erdős has proved that if such a $t$ exists then there must be infinitely many such $t$.

The sum-product problem. Erdős and Szemerédi [ErSz83] proved a lower bound of $\lvert A\rvert^{1+c}$ for some constant $c>0$, and an upper bound of $o(\lvert A\rvert^2)$. The lower bound has been improved a number of times. The current record is \[\max( \lvert A+A\rvert,\lvert AA\rvert)\gg\lvert A\rvert^{\frac{1558}{1167}-o(1)}\] due to Rudnev and Stevens [RuSt22] (note $1558/1167=1.33504\cdots$).

Asked by Erdős and Szemerédi [ErSz83]. Solved by Chang [Ch03].

The stated bounds are due to Burr and Erdős [BuEr85].

A paper of Burr and Erdős [BuEr85] proves both upper and lower bounds for $r=2$, showing that there exists some $c>0$ such that it cannot be true that \[\lvert A\cap \{1,\ldots,N\}\rvert \leq c(\log N)^2\] for all large $N$, and also constructing a Ramsey $2$-complete $A$ such that for all large $N$ \[\lvert A\cap \{1,\ldots,N\}\rvert \ll (\log N)^3.\] Burr has shown that the sequence of $k$th powers is Ramsey $r$-complete for every $r,k\geq 1$.

Solved by Conlon, Fox, and Pham [CFP21], who constructed for every $r\geq 2$ an $r$-Ramsey complete $A$ such that for all large $N$ \[\lvert A\cap \{1,\ldots,N\}\rvert \ll r(\log N)^2,\] and showed that this is best possible, in that there exists some constant $c>0$ such that if $A\subset \mathbb{N}$ satisfies \[\lvert A\cap \{1,\ldots,N\}\rvert \leq cr(\log N)^2\] for all large $N$ then $A$ cannot be $r$-Ramsey complete.

This was disproved for $k\geq 8$ by Khachatrian [AhKh94]. Erdős asks if the conjecture remains true provided $n\geq (1+o(1))p_k^2$.

A suitably constructed random set has this property if we are allowed to ignore an exceptional set of density zero. The challenge is obtaining this with no exceptional set. Erdős believed the answer should be no. Erdős and Sárkzözy proved that \[\frac{\lvert 1_A\ast 1_A(n)-\log n\rvert}{\sqrt{\log n}}\to 0\] is impossible. Erdős suggests it may even be true that the $\liminf$ and $\limsup$ of $1_A\ast 1_A(n)/\log n$ are always separated by some absolute constant.

Erdős [Er48] proved that $\sum_n \frac{d(n)}{2^n}$ is irrational, where $d(n)$ is the divisor function.

Erdős could prove this is true for almost all $n\equiv 0\pmod{4}$.

Conjectured by Erdős, Sós, and Sárkzözy [ErSaSo95], who proved this for $F_3(N)$. Erdős [Er35] earlier proved that $F_4(N)=o(N)$. Erdős also asks about $F(N)$, the size of the largest such set such that the product of no odd number of $a\in A$ is a square. Ruzsa proved that $\lim F(N)/N <1$. The value of $\lim F(N)/N$ is unknown, but it is $>1/2$.

Conjectured by Erdős, Pomerance, and Sárközy [ErPoSa97] prove this when $f$ is the divisor function or the number of distinct prime divisors of $n$, but Erdős believes it is false when $f(n)=\phi(n)$ or $\sigma(n)$.

Conjectured by Erdős and Lewin [ErLe96], who (among other related results) prove this when $a=3$, $b=5$, and $c=7$.

Conjectured by Burr, Erdős, Graham, and Li [BuErGrLi96]. Pomerance observed that the condition $\sum 1/(d_i-1)\geq 1$ is necessary. In [BuErGrLi96] they prove the property holds for $\{3,4,7\}$.

Conjectured by Burr, Erdős, Graham, and Li [BuErGrLi96].

Investigated by Erdős and Turán [ErTu34] in their first joint paper, where they proved that \[\log n \ll f(n) \ll n/\log n\] (the upper bound is trivial, taking $A=\{1,\ldots,n\}$). Erdős says that $f(n)=o(n/\log n)$ has never been proved, but perhaps never seriously attacked.

It is easy to see that we must have $\lvert A\rvert \ll N^{1/2}$. Csaba has constructed such an $A$ with $\lvert A\rvert \gg N^{1/5}$.

Erdős and Selfridge proved that $N$ can never be a perfect power. Erdős remarked that this 'seems hopeless at present'.

The answer is yes (in fact with $2$ replaced with any constant $c>1$), proved by Maier and Tenenbaum [MaTe84].

See also [449].

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 [Er35] proved that this sum always converges for a primitive set. Solved by Lichtman [Li23].

Proved by Sárkzözy [Sa85] for all sufficiently large $n$, and by Granville and Ramaré [GrRa96] for all $n\geq 5$.

More generally, if $f(n)$ is the largest integer such that, for some prime $p$, we have $p^{f(n)}$ dividing $\binom{2n}{n}$, then $f(n)$ should tend to infinity with $n$. Can one even disprove that $f(n)\gg \log n$?

It is easy to see by probabilistic methods that this holds for almost all integers. Romanoff [Ro34] showed that a positive density set of integers are representable as the sum of $2^k+p$ for prime $p$, and Erdős used covering systems to show that there is a positive density set of integers which cannot be so represented.

If $x$ is irrational then this can be false, but it may be true for almost all $x\in\mathbb{R}$. Curtiss [Cu22] showed that this is true for $x=1$ and Erdős [Er50b] showed it is true for all $x\in\mathbb{Q}$.

Erdős [Er51] showed that there are infinitely many $n$ such that \[s_{n+1}-s_n > (1+o(1))\frac{\pi^2}{6}\frac{\log s_n}{\log\log s_n},\] so this bound would be the best possible. The current best bound is due to Chan [Ch21], who proved an upper bound of $s_n^{\frac{5}{26}+o(1)}$. More precisely, they proved that there exists some $C>0$ such that, for all large $x$, the interval \[(x,x+Cx^{5/26}]\] contains a squarefree number.

See also [489].

The answer is yes, proved by Green and Tao [GrTa08]. The stronger claim that there are arbitrarily long arithmetic progressions of consecutive primes is still open.

The answer is yes, as proved by Montgomery and Vaughan [MoVa86], who in fact found the correct order of magnitude with the power $2$ replaced by any $\gamma\geq 1$.

Lorentz [Lo54] showed there is such a set with, for all large $N$, \[\lvert A\cap\{1,\ldots,N\}\rvert \ll \frac{\log\log N}{\log N}N\]

Erdős [Er51] proved that, for infinitely many $k$, \[ n_{k+1}-n_k \gg \frac{\log n_k}{\sqrt{\log\log n_k}}.\] Richards [Ri82] improved this to \[\limsup_{k\to \infty} \frac{n_{k+1}-n_k}{\log n_k} \geq 1/4.\] The constant $1/4$ here has been improved, most lately to $0.868\cdots$ by Dietmann, Elsholtz, Kalmynin, Konyagin, and Maynard [DEKKM22]. The best known upper bound is due to Bambah and Chowla [BaCh47], who proved that \[n_{k+1}-n_k \ll n_k^{1/4}.\]

Cramer proved an upper bound of $O(N(\log N)^4)$ conditional on the Riemann hypothesis. The prime number theorem immediately implies a lower bound of $\gg N(\log N)^2$.

Erdős [Er50] proved that there are infinitely many $n$ such that $f(n)\gg \log\log n$. Erdős could not even prove that there do not exist infinitely many integers $n$ such that for all $2^k<n$ the number $n-2^k$ is prime (probably $n=105$ is the largest such integer).

Erdős [Er50] proved this when $A=\{2^k : k\geq 0\}$. Solved by Chen and Ding [ChDi22].

This is well-known if $c_1$ is sufficiently small.

Wintner observed that if $f$ can take complex values on the unit circle then the limit need not exist. The answer is yes, as proved by Halász [Ha68].

Originally asked to Erdős by Wintner. The limit is infinite for a finite set of primes, which follows from a theorem of Pólya.

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

A sequence defined in such a fashion is known as Sylvester's sequence.

Originally asked to Erdős by Kalmár. Erdős believed the answer is yes. Romanoff [Ro34] proved that the answer is yes if $C$ is an integer.

Proved by Birch [Bi59].

Erdős [Er75c] proved the answer is yes under the stronger condition that $\limsup n_k/k^t=\infty$ for all $t\geq 1$.

Related to [69]. Erdős and Graham [ErGr80] write 'we just know too little about sieves to be able to handle such a question ("we" here means not just us but the collective wisdom (?) of our poor struggling human race).'

The answer is yes, as shown by Nesterenko [Ne96].

This is known now for $1\leq k\leq 4$. The cases $k=1,2$ are reasonably straightforward, as observed by Erdős [Er52]. The case $k=3$ was proved independently by Schlage-Puchta [ScPu06] and Friedlander, Luca, and Stoiciu [FLC07]. The case $k=4$ was proved by Pratt [Pr22].

This was disproved by Cassels [Ca60].

Cassels [Ca60] proved this under the alternative hypotheses \[\lvert A\cap [1,2x]\rvert -\lvert A\cap [1,x]\rvert\gg \log\log x\] and \[\sum_{n\in A} \{ \theta n\}^2=\infty\] for every $\theta\in (0,1)$.

Related to [260].

Erdős and Straus proved the answer is yes for the 2-dimensional version, where $X\subseteq \mathbb{R}^2$ is the set of \[\left( \sum_{n\in A} \frac{1}{n},\sum_{n\in A}\frac{1}{n+1}\right) \] as $A\subseteq\mathbb{N}$ ranges over all infinite sets with $\sum_{n\in A}\frac{1}{n}<\infty$.

Selfridge has found an example using divisors of $360$ if $p=3$ is allowed.

This is best possible as the system $2^{i-1}\pmod{2^i}$ shows. This was proved indepedently by Selfridge and Crittenden and Vanden Eynden [CrVE70].

Whether such a composite Lucas sequence even exists was open for a while, but using covering systems Graham [Gr64] showed that \[a_0 = 1786772701928802632268715130455793\] \[a_1 = 1059683225053915111058165141686995\] generate such a sequence. This problem asks whether one can have a composite Lucas sequence without 'an underlying system of covering congruences responsible'.

The density of such $n$ is zero.

Even the case $k=3$ seems difficult. This may be true with the primes replaced by any set $A\subseteq \mathbb{N}$ such that \[\lvert A\cap [1,N]\rvert \gg N/\log N\] and \[\sum_{\substack{n\in A\\ n\leq N}}\frac{1}{n} -\log\log N\to \infty\] as $N\to \infty$.

For $k=1$ or $k=2$ any set $A$ such that $\sum_{n\in A}\frac{1}{n}=\infty$ has this property.

Erdős and Graham [ErGr80] suggest this 'may have to await improvements in current sieve methods' (of which there have been several since 1980).

The latter condition is clearly sufficient, the problem is if it's also necessary. The assumption implies $\sum \frac{1}{n_i}=\infty$.

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 $n_i=n_j$.

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.

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

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

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

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 [Bl23] (coupled with the trivial greedy approach) implies that $k(N)=(1-o(1))\log N$.

It is not even known whether this is $2^{cN}$ for some $c<1$ or $2^{(1+o(1))N}$.

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

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

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

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.

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 \[D(p) \gg p\log p.\]

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).

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 $Q$?

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 $\{1,\ldots,N\}$.

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

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

For example, \[\frac{1}{2}+\frac{1}{3}=1-\frac{1}{6}.\]

$c_0$ is called the Vardi constant.

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 [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 \[\sum_{k\geq 2}\frac{1}{k^2}<2.\]

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].

Connected to Waring's problem. The famous Hypothesis $K$ of Hardy and Littlewood was that $1_A^{(k)}(n)\leq n^{o(1)}$, but this was disproved by Mahler [Ma36] for $k=3$, who constructed infinitely many $n$ such that \[1_A^{(3)}(n)\gg n^{1/12}\] (where $A$ is the set of cubes). Erdős believed Hypothesis $K$ fails for all $k\geq 4$, but this is unknown. Hardy and Littlewood made the weaker Hypothesis $K^*$ that for all $N$ and $\epsilon>0$ \[\sum_{n\leq N}1_A^{(k)}(n)^2 \ll_\epsilon N^{1+\epsilon}.\] Erdős and Graham remark: 'This is probably true but no doubt very deep. However, it would suffice for most applications.'

Independently Erdős [Er36] and Chowla proved that for all $k\geq 3$ and infinitely many $n$ \[1_A^{(k)}(n) \gg n^{c/\log\log n}\] for some constant $c>0$ (depending on $k$).

This would have significant applications to Waring's problem. Erdős and Graham describe this as 'unattackable by the methods at our disposal'. The case $k=2$ was resolved by Landau, who showed \[f_{2,2}(x) \sim \frac{cx}{\sqrt{\log x}}\] for some constant $c>0$.

For $k>2$ it is not known if $f_{k,k}(x)=o(x)$.

Erdős and Graham describe this problem as 'very annoying'. Probably $f(x)=x^5$ should work.

Mahler and Erdős [ErMa38] proved that $f_{k,2}(x) \gg x^{2/k}$. For $k=3$ the best known is due to Wooley [Wo15], \[f_{3,3}(x) \gg x^{0.917\cdots}.\]

Erdős originally asked whether this was true with $A=B$, but this was disproved by Cassels [Ca57].

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

Asked by Erdős and Newman. Nešetřil and Rödl have shown the answer is no for all $C$ (source is cited as 'personal communication' in [ErGr80]). Erdős had previously shown the answer is no for $C=3,4$ and infinitely many other values of $C$.

Erdős proved that $1/2$ is possible and Krückeberg [Kr61] proved $1/\sqrt{2}$ is possible. Erdős and Turán [ErTu41] have proved this $\limsup$ is always $\leq 1$.

The fact that $1$ is possible would follow if any finite Sidon set is a subset of a perfect difference set.

Asked by Erdős and Nathanson.

Prikry, Tijdeman, Stewart, and others (see the survey articles [St78] and [Ti79]) have shown that a sufficient condition is that $A$ has positive density.

One can also ask what conditions are sufficient for $D(A)$ to have positive density, or for $\sum_{d\in D(A)}\frac{1}{d}=\infty$, or even just $D(A)\neq\emptyset$.

Erdős and Newman [ErNe77] have proved this is true when $A$ is the set of squares.

Erdős originally asked if even $f(n)\leq n^{1/3}$ is true. This is known, and the best bound is due to Balog [Ba89] who proved that \[f(n) \ll_\epsilon n^{\frac{4}{9\sqrt{e}}+\epsilon}\] for all $\epsilon>0$. (Note $\frac{4}{9\sqrt{e}}=0.2695\ldots$.)

It is likely that $f(n)\leq n^{o(1)}$, or even $f(n)\leq e^{O(\sqrt{\log n})}$.

One way this can happen is if there exists $\theta>0$ such that \[A=\{ n>0 : \{ n\theta\} \in X_A\}\textrm{ and }B=\{ n>0 : \{n\theta\} \in X_B\}\] where $\{x\}$ denotes the fractional part of $x$ and $X_A,X_B\subseteq \mathbb{R}/\mathbb{Z}$ are such that $\mu(X_A+X_B)=\mu(X_A)+\mu(X_B)$. Are all possible $A$ and $B$ generated in a similar way (using other groups)?

Erdős and Graham [ErGr80b] have shown that a basis $A$ has an exact order if and only if $a_2-a_1,a_3-a_2,a_4-a_3,\ldots$ are coprime. They also prove that \[\frac{1}{4}\leq \lim_r \frac{h(r)}{r^2}\leq \frac{5}{4}.\] It is known that $h(2)=4$, but even $h(3)$ is unknown (it is $\geq 7$).

Bateman has observed that for $h\geq 3$ there is a basis of order $h$ with no restricted order, taking \[A=\{1\}\cup \{x>0 : h\mid x\}.\] Kelly [Ke57] has shown that any basis of order $2$ has restricted order at most $4$ and conjectures it always has restricted order at most $3$.

The set of squares has order $4$ and restricted order $5$ (see [Pa33]) and the set of triangular numbers has order $3$ and restricted order $3$ (see [Sc54]).

Is it true that if $A\backslash F$ is a basis for all finite sets $F$ then $A$ must have a restricted order? What if they are all bases of the same order?

Erdős and Graham [ErGr80] also ask about the difference set $A-A$, whether this has positive density, and whether this contains $22$.

This sequence is at OEIS A005282.

An old problem of Dickson. Even a starting set as small as $\{1,4,9,16,25\}$ requires thousands of terms before periodicity occurs.

A problem of Ulam. The sequence is \[1,2,3,4,6,8,11,13,16,18,26,28,\ldots\] at OEIS A002858.

A problem of Folkman. Folkman [Fo66] showed that this is true if \[\lvert A\cap \{1,\ldots,N\}\rvert\gg N^{1+\epsilon}\] for some $\epsilon>0$ and all $N$.

The original question was answered by Szemerédi and Vu [SzVu06] (who proved that the answer is yes).

This is best possible, since Folkman [Fo66] showed that for all $\epsilon>0$ there exists a multiset $A$ with \[\lvert A\cap \{1,\ldots,N\}\rvert\ll N^{1+\epsilon}\] for all $N$, such that $A$ is not subcomplete.

Folkman proved this under the stronger assumption that \[\lvert A\cap \{1,\ldots,N\}\rvert\gg N^{1/2+\epsilon}\] for some $\epsilon>0$.

This is true, and was proved by Szemerédi and Vu [SzVu06]. The stronger conjecture that this is true under \[\lvert A\cap \{1,\ldots,N\}\rvert\geq (2N)^{1/2}\] seems to be still open (this would be best possible as shown by [Er61b].

Erdős and Graham [ErGr80] remark that very little is known about $T(A)$ in general. It is known that \[T(n)=1, T(n^2)=128, T(n^3)=12758,\] \[T(n^4)=5134240,\textrm{ and }T(n^5)=67898771.\] Erdős and Graham remark that a good candidate for the $n$ in the question are $k=2^t$ for large $t$, perhaps even $t=3$, because of the highly restricted values of $n^{2^t}$ modulo $2^{t+1}$.

Graham [Gr64d] has shown that the sequence $a_n=F_n-(-1)^{n}$, where $F_n$ is the $n$th Fibonacci number, has these properties. Erdős and Graham [ErGr80] remark that it is easy to see that if $a_{n+1}/a_n>\frac{1+\sqrt{5}}{2}$ then the second property is automatically satisfied, and that it is not hard to construct very irregular sequences satisfying both properties.

The Fibonacci sequence $1,1,2,3,5,\ldots$ shows that $m=1$ and $n=2$ is possible. The sequence of powers of $2$ shows that $m=0$ and $n=1$ is possible. The case $m=2$ and $n=3$ is not known.

Even in the range $t\in (0,1)$ and $\alpha\in (1,2)$ the behaviour is surprisingly complex. For example, Graham [Gr64e] has shown that for any $k$ there exists some $t_k\in (0,1)$ such that the set of $\alpha$ such that the sequence is complete consists of at least $k$ disjoint line segments. It seems likely that the sequence is complete for all $t>0$ and all $1<\alpha < \frac{1+\sqrt{5}}{2}$. Proving this seems very difficult, since we do not even known whether $\lfloor (3/2)^n\rfloor$ is odd or even infinitely often.

This was proved by Ryavec. The stronger statement that, for all $s\geq 0$, \[\sum_{n\in A}\frac{1}{n^s} <\frac{1}{1-2^{-s}},\] was proved by Hanson, Steele, and Stenger [HSS77].

Graham [Gr64f] proved this is true when $p(n)=n$. Erdős and Graham also ask which rational functions $r(x)\in\mathbb{Z}[x]$ force $\{ r(n) : n\in\mathbb{N}\}$ to be complete?

Bleicher and Erdős conjecture the answer is no.

This fails for $a_i=i$ for example. Erdős and Graham also ask what happens if we drop the monotonicity restriction and just ask that the $a_i$ are distinct. Perhaps some permutation of $\{1,\ldots,n\}$ has at least $cn^2$ such distinct sums. This latter problem has been solved by Konieczny [Ko15], who showed that a random permutation of $\{1,\ldots,n\}$ has, with high probability, at least \[\left(\frac{1+e^{-2}}{4}+o(1)\right) n^2\] many such consecutive sums.

The original problem was solved (in the affirmative) by Beker [Be23b].

They also ask how many consecutive integers $>n$ can be represented as such a sum? Is it true that, for any $c>0$ at least $cn$ such integers are possible (for sufficiently large $n$)?

Asked by Erdős and Harzheim. What if we remove the monotonicity and/or the distinctness constraint? Also what is the least $m$ which is not a sum of the given form? Can it be much larger than $n$? Erdős and Harzheim can show that $\sum_{x<a_i<x^2}\frac{1}{a_i}\ll 1$. Is it true that $\sum_i \frac{1}{a_i}\ll 1$?

Erdős and Moser [Mo63] considered the case when $A$ is the set of primes, and conjectured that the $\limsup$ of the number of such representations in this case is infinite. They could not even prove that the upper density of the set of integers representable in this form is positive.

A problem of MacMahon, studied by Andrews [An75].

Erdős and Graham state it 'can be shown' that $f(n)\to \infty$.

Erdős and Moser proved this bound with an additional factor of $(\log n)^{3/2}$. This was removed by Sárközy and Szemerédi [SaSz65]. Stanley [St80] has shown that this quantity is maximised when $A$ is an arithmetic progression and $t=\tfrac{1}{2}\sum_{n\in A}n$.

Erdős and Selfridge have proved that the product of consecutive integers is never a power. The condition $\lvert I_i\rvert \geq 4$ is necessary here, since Pomerance has observed that the product of \[(2^{n-1}-1)2^{n-1}(2^{n-1}+1),\] \[(2^n-1)2^n(2^n+1),\] \[(2^{2n-1}-2)(2^{2n-1}-1)2^{2n-1},\] and \[(2^{2n-2}-2)(2^{2n}-1)2^{2n}\] is a always a square.

Erdős originally asked Mahler whether there are infinitely many pairs of consecutive powerful numbers, but Mahler immediately observed that the answer is yes from the infinitely many solutions to the Pell equation $x^2=8y^2+1$.

Erdős originally asked Mahler whether there are infinitely many pairs of consecutive powerful numbers, but Mahler immediately observed that the answer is yes from the infinitely many solutions to the Pell equation $x^2=8y^2+1$.

Erdős originally asked Mahler whether there are infinitely many pairs of consecutive powerful numbers, but Mahler immediately observed that the answer is yes from the infinitely many solutions to the Pell equation $x^2=8y^2+1$.

Note that $8$ is 3-full and $9$ is 2-full. Is the the only pair of such consecutive integers?

It would also be interesting to find upper and lower bounds for the analogous product with $B_r$ for $r\geq 3$, where $B_r(n)$ is the $r$-full part of $n$ (that is, the product of prime powers $p^a \mid n$ such that $p^{a+1}\nmid n$ and $a\geq r$). Is it true that, for every fixed $r,k\geq 2$ and $\epsilon>0$, \[\limsup \frac{\prod_{n\leq m<n+k}B_r(m) }{n^{1+\epsilon}}\to\infty?\]

Mahler [Ma35] showed that this is $\gg \log\log n$. Schinzel [Sc67b] observed that for infinitely many $n$ it is $\leq n^{O(1/\log\log\log n)}$. The truth is probably $\gg (\log n)^2$ for all $n$.

Open even for $k=2$. Erdős and Graham remark 'the answer should be affirmative but the problem seems very hard'.

Pomerance has observed that if we replace $1/2$ in the exponent by $1/\sqrt{e}-\epsilon$ for any $\epsilon>0$ then this is true for density reasons.

Conjectured by Erdős and Pomerance [ErPo78], who proved that this set and its complement both have positive upper density.

Conjectured by Erdős and Pomerance [ErPo78], who proved the analogous result for $P(n)<P(n+1)<P(n+2)$. Proved by Balog [Ba01], who proved that this is true for $\gg \sqrt{x}$ many $n\leq x$ (for all large $x$).

This would follow if $P(n(n+1))/\log n\to \infty$, where $P(m)$ denotes the largest prime factor of $m$ (see Problem [368]). Hickerson conjectured the largest solution is \[16! = 14! 5!2!.\] The condition $a_1<n-1$ is necessary to rule out the trivial solutions when $n=a_2!\cdots a_k!$.

Studied by Erdős and Graham [ErGr76] (see also [LSS14]). It is known, for example, that:

• no $D_k$ contains a prime,
• $D_2=\{ n^2 : n>1\}$,
• $\lvert D_3\cap \{1,\ldots,n\}\rvert = o(\lvert D_4\cap \{1,\ldots,n\}\rvert)$,
• the least element of $D_6$ is $527$, and
• $D_k=\emptyset$ for $k>6$.

Very deep if true, since it would imply there is a prime between any two consecutive squares.

Originally conjectured by Grimm [Gr69], who proved it when $k \ll \log n/\log\log n$. Ramachandra, Shorey, and Tijdeman [RST75] have improved this to \[k \ll \left(\frac{\log n}{\log\log n}\right)^3.\]

Erdős, Graham, Ruzsa, and Straus [EGRS75] have shown that, for any two primes $p$ and $q$, there are infinitely many $n$ such that $\binom{2n}{n}$ is coprime to $pq$.

If the sum is denoted by $f(n)$ then Erdős, Graham, Ruzsa, and Straus [EGRS75] have shown \[\lim_{x\to \infty}\frac{1}{x}\sum_{n\leq x}f(n) = \sum_{k=2}^\infty \frac{\log k}{2^k}=\gamma_0\] and \[\lim_{x\to \infty}\frac{1}{x}\sum_{n\leq x}f(n)^2 = \gamma_0^2,\] so that for almost all integers $f(m)=\gamma_0+o(1)$.

Erdős and Graham state they can prove that, for $k$ fixed and large, the density of $n$ such that $\binom{n}{k}$ is squarefree is $o_k(1)$. They can also prove that there are infinitely many $n$ such that $\binom{n}{k}$ is not squarefree for $1\leq k<n$, and expect that the density of such $n$ is positive.

If $s(n)$ denotes the largest integer such that $\binom{n}{k}$ is divisible by $p^{s(n)}$ for some prime $p$ for at least one $1\leq k<n$ then it is easy to see that $s(n)\to \infty$ as $n\to \infty$ (and in fact that $s(n) \asymp \log n$).

Erdős and Graham only knew that $B(x) > x^{1-o(1)}$. Similarly, we call an interval $[u,v]$ 'very bad' if $\prod_{u\leq m\leq v}m$ is powerful. The number of integers $n\leq x$ contained in at least one very bad interval should be $\ll x^{1/2}$. In fact, it should be asymptotic to the number of powerful numbers $\leq x$.

See also [382].

Conjectured by Erdős and Selfridge. There are infinitely many $n$ such that $n(n+1)$ is powerful (see [364]).

Erdős and Graham report it follows from results of Ramachandra that $v-u\leq v^{1/2+o(1)}$.

See also [380].

A conjecture of Erdős and Selfridge. Proved by Ecklund [Ec69], who made the stronger conjecture that whenever $n>k^2$ the binomial coefficient $\binom{n}{k}$ is divisible by a prime $p<n/k$. They have proved the weaker inequality $p\ll n/k^c$ for some constant $c>0$.

Erdős and Graham write that 'a proof that this cannot happen infinitely often for $\binom{n}{2}$ seems hopeless; probably this can never happen for $\binom{n}{k}$ if $3\leq k\leq n-3$.'

Erdős once conjectured that $\binom{n}{k}$ must always have a divisor in $(n-k,n]$, but this was disproved by Schinzel and Erdős [Sc58].

Asked by Erdős and Straus. For example when $n=2$ we have $k=4$: \[3\cdot 4\cdot 5\cdot 6 \mid 7\cdot 8\cdot 9\cdot 10.\] No example was known for $n\geq 10$.

Erdős, Guy, and Selfridge [EGS82] have shown that \[f(n)-2n \asymp \frac{n}{\log n}.\]

Erdős, Selfridge, and Straus have shown that \[\lim \frac{t(n)}{n}=\frac{1}{e}.\] Alladi and Grinstead [AlGr77] have obtained similar results when the $a_i$ are restricted to prime powers.

If we change the condition to $a_t\leq n$ it can be shown that \[A(n)=n-\frac{n}{\log n}+o\left(\frac{n}{\log n}\right)\] via a greedy decomposition (use $n$ as often as possible, then $n-1$, and so on). Other questions can be asked for other restrictions on the sizes of the $a_t$.

Erdős and Graham write that they do not even know whether $f(n)=1$ infinitely often (i.e. whether a factorial is the product of two consecutive integers infinitely often).

The answer is yes, proved by Hall. It is probably true that the sum is $o(N/(\log N)^c)$ for some constant $c>0$.

It follows from the Thue-Siegel theorem that, for $n$ fixed, there are only finitely many potential such $x$ and $y$.

Erdős and Graham write that $n+1$ always divides $\binom{2n}{n}$, but it is quite rare that $n$ divides $\binom{2n}{n}$.

The Brocard-Ramanujan conjecture. Erdős and Graham describe this as an old conjecture, and write it 'is almost certainly true but it is intractable at present'.

Overholt [Ov93] has shown that this has only finitely many solutions assuming a weak form of the abc conjecture.

Erdős and Obláth [ErOb37] proved this is true when $(x,y)=1$ and $k\neq 4$. Pollack and Shapiro [PoSh73] proved this is true when $(x,y)=1$ and $k=4$. The known methods break down without the condition $(x,y)=1$.

Erdős and Graham write that it is easy to show that $g_k(n) \ll_k \log n$ always, but the best possible constant is unknown.

See also [401].

See also [400].

A conjecture of Graham [Gr70], who also conjectured that (assuming $A$ itself has no common divisor) then the only cases where equality achieved are when $A=\{1,\ldots,n\}$ or $\{L/1,\ldots,L/n\}$ (where $L=\mathrm{lcm}(1,\ldots,n)$) or $\{2,3,4,6\}$.

Proved for all sufficiently large sets (including the sharper version which characterises the case of equality) independently by Szegedy [Sz86] and Zaharescu [Za87].

Proved for all sets by Balasubramanian and Soundararajan [BaSo96].

Asked by Burr and Erdős. Frankl and Lin [Li76] independently showed that the answer is yes, and the largest solution is \[2^7=2!+3!+5!.\] In fact Lin showed that the largest power of $2$ which can divide a sum of distinct factorials containing $2$ is $2^{254}$, and that there are only 5 solutions to $3^m=a_1!+\cdots+a_k!$ (when $m=0,1,2,3,6$).

See also [404].

See also [403]. Lin [Li76] has shown that $f(2,2) \leq 254$.

Erdős and Graham remark that it is probably true that in general $(p-1)!+a^{p-1}$ is rarely a power at all (although this can happen, for example $6!+2^6=28^2$).

The only examples seem to be $4=1+3$ and $256=1+3+3^2+3^5$. If we only allow the digits $1$ and $2$ then $2^{15}$ seems to be the largest such power of $2$.

A conjecture originally due to Newman.

Pillai [Pi29] was the first to investigate this function, and proved \[\log_3 n < f(n) < \log_2 n\] for all large $n$. Shapiro [Sh50] proved that $f(n)$ is essentially multiplicative.

Erdős, Granville, Pomerance, and Spiro [EGPS90] have proved that the answer to the first two questions is yes, conditional on a form of the Elliott-Halberstam conjecture.

It is likely true that, if $k\to \infty$ however slowly with $n$, then for almost $n$ the largest prime factor of $n$ is $\leq n^{o(1)}$.

A problem of Finucane. One can also ask about $n\mapsto \sigma(n)-1$.

The known solutions to $g_{k+2}(n)=2g_k(n)$ are $n=10$ and $n=94$. Selfridge and Weintraub found solutions to $g_{k+9}(n)=9g_k(n)$ and Weintraub found \[g_{k+25}(3114)=729g_k(3114)\] for all $k\geq 6$.

That is, there is (eventually) only one possible sequence that the iterated sum of divisors function can settle on. Selfridge reports numerical evidence suggests the answer is no, but Erdős and Graham write 'it seems unlikely that anything can be proved about this in the near future'.

See also [413] and [414].

Erdős suggested this problem as a way of showing that the iterated behaviour of $n\mapsto n+\nu(n)$ eventually settles into a single sequence, regardless of the starting value of $n$ (see also [412] and [414]).

Erdős and Graham report it could be attacked by sieve methods, but 'at present these methods are not strong enough'. Can one show that there exists an $\epsilon>0$ such that there are infinitely many $n$ where $m+\epsilon \nu(m)\leq n$ for all $m<n$?

Asked by Spiro. That is, there is (eventually) only one possible sequence that the iterations of $n\mapsto h(n)$ can settle on. Erdős and Graham believed the answer is yes. Similar questions can be asked by the iterates of many other functions. See also [412] and [413].

Erdős [Er36b] proved that \[F(n)\asymp \log\log\log n,\] and similarly if we replace $\phi$ with $\sigma$ or $\tau$ or $\nu$ or any 'decent' additive or multiplicative function.

Pillai [Pi29] proved $V(x)=o(x)$. The behaviour of $V(x)$ is almost completely understood. Maier and Pomerance [MaPo88] proved \[V(x)=\frac{x}{\log x}e^{(C+o(1))\log\log\log x},\] for some explicit constant $C>0$. Ford [Fo98] improved this to \[V(x)\asymp\frac{x}{\log x}e^{C_1(\log\log\log x-\log\log\log\log x)^2+C_2\log\log\log x-C_3\log\log\log\log x}\] for some explicit constants $C_1,C_2,C_3>0$. Unfortunately this falls just short of an asymptotic formula for $V(x)$ and determining whether $V(2x)/V(x)\to 2$.

See also [417].

It is trivial that $V'(x) \leq V(x)$. See also [416].

Asked by Erdős and Sierpiński. It follows from the Goldbach conjecture that every odd number can be written as $n-\phi(n)$. What happens for even numbers?

Erdős [Er73b] has shown that a positive density set of integers cannot be written as $\sigma(n)-n$.

It can be shown that any number of the shape $1+1/k$ for $k\geq 1$ is a limit point, but no others are known.

Erdős and Graham write that it is easy to show that $\lim F(n^{1/2},n)=\infty$, and in fact the $n^{1/2}$ can be replaced by $n^{1/2-c}$ for some small constant $c>0$.

Asked by Hofstadter. The sequence begins $1,1,2,3,3,4,\ldots$ and is A005185 in the OEIS. It is not even known whether $f(n)$ is well-defined for all $n$.

Asked by Hofstadter. The sequence begins $1,2,3,5,6,8,10,11,\ldots$ and is A005243 in the OEIS.

Asked by Hofstadter. The sequence begins $2,3,5,9,14,17,26,\ldots$ and is A005244 in the OEIS.

Erdős [Er68] proved that \[F(n)-\pi(n)\asymp n^{3/4}(\log n)^{-3/2}.\] Surprisingly, if we consider the corresponding problem in the reals (so the largest $m$ such that there are reals $1\leq a_1<\cdots <a_m\leq x$ such that for any distinct indices $i,j,r,s$ we have $\lvert a_ia_j-a_ra_s\rvert \geq 1$) then Alexander proved that $m> x/8e$ is possible.

See also [490].

The answer is yes, proved by Ruzsa [Ru72].

Erdős and Graham could show this is true (assuming the prime $k$-tuple conjecture) if we replace $\liminf$ by $\limsup$.

Erdős and Graham write 'preliminary calculations made by Selfridge indicate that this is the case but no proof is in sight'. For example if $n=8$ we have $a_1=7$ and $a_2=5$ and then must stop.

A problem of Ostmann, sometimes known as the 'inverse Goldbach problem'. The answer is surely no. The best result in this direction is due to Elsholtz and Harper [ElHa15], who showed that if $A,B$ are such sets then for all large $x$ we must have \[\frac{x^{1/2}}{\log x\log\log x} \ll \lvert A \cap [1,x]\rvert \ll x^{1/2}\log\log x\] and similarly for $B$.

Elsholtz [El01] has proved there are no infinite sets $A,B,C$ such that $A+B+C$ agrees with the set of prime numbers up to finitely many exceptions.

See also [432].

Asked by Straus, inspired by a problem of Ostmann (see [431]).

This type of problem is associated with Frobenius. Erdős and Graham [ErGr72] proved $g(n,t)<2t^2/n$, and there are examples which show that \[g(n,t) \geq \frac{t^2}{n-1}-5t\] for $n\geq 2$.

Associated with problems of Frobenius.

Asked by Lehmer and Lehmer [LeLe62]. For example $\Lambda(2,2)=9$ and $\Lambda(3,2)=77$. It is known that $\Lambda(k,3)=\infty$ for all even $k$ and $\Lambda(k,4)=\infty$ for all $k$.

This has been partially resolved: Hildebrand [Hi91] has shown that $\Lambda(k,2)$ is finite for all $k$.

There are trivially $o(x)$ many squares.

Taking all integers $\equiv 1\pmod{3}$ shows that $\lvert A\rvert\geq N/3$ is possible. This can be improved to $\tfrac{11}{32}N$ by taking all integers $\equiv 1,5,9,13,14,17,21,25,26,29,30\pmod{32}$, as observed by Massias.

Lagarias, Odlyzko, and Shearer [LOS83] proved this is sharp for the modular version of the problem; that is, if $A\subseteq \mathbb{Z}/N\mathbb{Z}$ is such that $A+A$ contains no squares then $\lvert A\rvert\leq \tfrac{11}{32}N$. They also prove the general upper bound of $\lvert A\rvert\leq 0.475N$ for the integer problem.

In fact $\frac{11}{32}$ is sharp in general, as shown by Khalfalah, Lodha, and Szemerédi [KLS02], who proved that the maximal such $A$ satisfies $\lvert A\rvert\leq (\tfrac{11}{32}+o(1))N$.

It is easy to give a sequence with \[\limsup\frac{A(x)}{x^{1/2}}=c>0.\] There are related results (particularly for what happens if we consider $\mathrm{lcm}(a_i,a_{i+1},\ldots,a_{i+k})$ in a paper of Erdős and Szemerédi [ErSz80].

The answer is yes, proved by Erdős and Sárkőzy [ErSa80].

An unpublished result of Heilbronn guarantees this for $c$ sufficiently close to $1$.

Erdős [Er35] proved that $\delta(n)<(\log n)^{-c}$ for some $c>0$. Tenenbaum [Te75] showed that $\delta(n)=(\log n)^{(-1+o(1))\alpha}$ where \[\alpha=1-\frac{1+\log\log 2}{\log 2}=0.08607\cdots.\]

This has been resolved by Ford [Fo08]. Among many other results, Ford proves \[\delta(n)\asymp \frac{1}{(\log n)^\alpha(\log\log n)^{3/2}},\] and that the second conjecture is false (i.e. $\delta'(n) \geq \delta(n)$ for some constant $c>0$)

Erdős and Graham also ask whether there is a good inequality known for $\sum_{n\leq x}\tau^*(n)$.

See also [447].

Erdős and Graham write 'we can prove $n_k>k^{1+c}$ but no doubt much more is true'.

Erdős [Er37] proved that the density of integers $n$ with $\omega(n)>\log\log n$ is $1/2$. The Chinese remainder theorem implies that there is such an interval with \[\lvert I\rvert \geq (1+o(1))\frac{\log x}{(\log\log x)^2}.\] It could be true that there is such an interval of length $(\log x)^{k}$ for arbitrarily large $k$.

Asked by Erdős and Straus. The answer is no, as shown by Pomerance [Po79].

Pomerance [Po79] has proved the $\limsup$ is at least $2$.

Richter [Ri76] proved that \[\liminf_n \frac{q_n}{n^2}>2.84\cdots.\]

Erdős and Graham write this is 'almost certainly' true, but the proof is beyond our ability, for two reasons (at least):

• Firstly, one has to rule out the possibility of many primes $q$ such that $p_k<q^2<p_{k+1}$. There should be at most one such $q$, which would follow from $p_{k+1}-p_k<p_k^{1/2}$, which is essentially the notorious Legendre's conjecture.
• The small primes also cause trouble.

This question arose in work of Eggleton, Erdős, and Selfridge.

Erdős and Graham report they can show \[f(n,t) \gg \frac{t}{\log t}.\]

Erdős and Graham write the existence of such an $\alpha$ has 'very recently been shown', but frustratingly give neither a name nor a reference. I will try to track down this solution soon.

The first conjecture was proved by Sárközy [Sa76], who in fact proved \[N(X,\delta) \ll \delta^{-3}\frac{X}{\log\log X}.\] It is not even known whether $N(X,\delta)<X^{1-\epsilon}$ for some $\epsilon>0$. See also [466].

Graham proved this is true, and in fact \[N(X,1/10)> \frac{\log X}{10}.\] This was substantially improved by Sárközy [Sa76], who proved that for, all sufficiently small $\delta>0$, \[N(X,\delta)>X^{1/2-\delta^{1/7}}.\] See also [465].

This is what I assume the intended problem is, although the presentation in [ErGr80] is missing some crucial quantifiers, so I may have misinterpreted it.

The same question can be asked for those $n$ which do not have distinct sums of sets of divisors, but any proper divisor of $n$ does.

Weird numbers were investigated by Benkoski and Erdős [BeEr74], who proved that the set of weird numbers has positive density.

Melfi [Me15] has proved that there are infinitely many primitive weird numbers, conditional on various well-known conjectures on the distribution of prime gaps. For example, it would suffice to show that $p_{n+1}-p_n <\frac{1}{10}p_n^{1/2}$ for sufficiently large $n$.

A problem of Ulam. In particular, what about $Q=\{3,5,7,11\}$?

A problem due to Ulam. For example if we begin with $3,5$ then the sequence continues $3,5,7,11,13,17,\ldots$. It is possible that this sequence is infinite.

Asked by Segal. The answer is yes, as shown by Odlyzko.

Watts has suggested that perhaps the obvious greedy algorithm defines such a permutation - that is, let $a_1=1$ and let \[a_{n+1}=\min \{ x : a_n+x\textrm{ is prime and }x\neq a_i\textrm{ for }i\leq n\}.\] In other words, do all positive integers occur as some such $a_n$? Do all primes occur as a sum?

The sequence is OEIS A002048 and begins \[1,2,4,5,8,10,14,15,16,21,22,23,25,\ldots.\] In general, if $a_1<a_2<\cdots$ is a sequence so that no $a_n$ is a sum of consecutive $a_i$s, then must the density of the $a_i$ be zero? What about the lower density?

A problem of Graham.

A question of Erdős and Heilbronn. Solved in the affirmative by da Silva and Hamidoune [dSHa94].

Probably there is no such $A$ for any polynomial $f$.

A conjecture of Graham. It is easy to see that $2^n\not\equiv 1\mod{n}$ for all $n>1$, so the restriction $k\neq 1$ is necessary. Erdős and Graham report that Graham, Lehmer, and Lehmer have proved this for $k=2^i$ for $i\geq 1$, or if $k=-1$, but I cannot find such a paper.

As an indication of the difficulty, when $k=3$ the smallest $n$ such that $2^n\equiv 3\pmod{n}$ is $n=4700063497$.

A conjecture of Newman. This was proved Chung and Graham, who in fact show that for any $\epsilon>0$ there must exist some $n$ such that there are infinitely many $m$ for which \[\lvert x_{m+n}-x_m\rvert < \frac{1}{(c-\epsilon)n}\] where \[c=1+\sum_{k\geq 1}\frac{1}{F_{2k}}=2.535\cdots\] and $F_m$ is the $m$th Fibonacci number. This constant is best possible.

Erdős and Graham write that 'it is surprising that [this problem] offers difficulty'.

The result for $\sqrt{2}$ was obtained by Graham and Pollak [GrPo70]. The problem statement is open-ended, but presumably Erdős and Graham would have been satisfied with the wide-ranging generalisations of Stoll ([St05] and [St06]).

Schur proved that $f(k)<ek!$. See also [183].

A conjecture of Roth.

Davenport and Erdős [DaEr37] proved that the answer is yes when $X_n=\{0\}$ for all $n\in A$. The problem considers logarithmic density since Besicovitch [Be34] showed examples exist without a natural density, even when $X_n=\{0\}$ for all $n\in A$.

Davenport and Erdős [DaEr37] showed that there must exist an infinite sequence $a_1<a_2\cdots$ in $A$ such that $a_i\mid a_j$ for all $i\leq j$.

This is true, a consequence of the positive solution to [447] by Kleitman [Kl71].

The example $A=\{a\}$ and $n=2a-1$ and $m=2a$ shows that $2$ would be best possible.

For example, when $A=\{p^2: p\textrm{ prime}\}$ then $B$ is the set of squarefree numbers, and the existence of this limit was proved by Erdős.

See also [208].

This would be best possible, for example letting $A=[1,N/2]\cap \mathbb{N}$ and $B=\{ N/2<p\leq N: p\textrm{ prime}\}$.

See also [425].

Erdős [Er46] proved that if $f(n+1)-f(n)=o(1)$ or $f(n+1)\geq f(n)$ then $f(n)=c\log n$ for some constant $c$.

This is true, and was proved by Wirsing [Wi70].

For example if $A=\mathbb{N}$ then $f(x)=\{x\}$ is the usual fractional part operator.

A problem due to Le Veque [LV53], who proved it in some special cases.

This is false is general, as shown by Schmidt [Sc69].

A conjecture of Schinzel.

The infamous Littlewood conjecture.

Originally a conjecture due to Oppenheim. Davenport and Heilbronn [DaHe46] solve the analogous problem for quadratic forms in 5 variables.

This is true, and was proved by Margulis [Ma89].