OPEN

Let $A$ be the set of all integers not of the form $p+2^{k}+2^l$ (where $k,l\geq 0$ and $p$ is prime). Is the upper density of $A$ positive?

Crocker [Cr71] has proved there are are $\gg\log\log N$ such integers in $\{1,\ldots,N\}$ (any number of the form $2^{2^m}-1$ with $m\geq 3$ suffices). Pan [Pa11] improved this to $\gg_\epsilon N^{1-\epsilon}$ for any $\epsilon>0$. 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.

OPEN

Is there some $k$ such that every integer is the sum of a prime and at most $k$ powers of 2?

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 $1-\epsilon$.

Granville and Soundararajan [GrSo98] have conjectured that at most $3$ powers of 2 suffice for all odd integers, and hence at most $4$ powers of $2$ suffice for all even integers. (The restriction to odd integers is important here - for example, Bogdan Grechuk has observed that $1117175146$ is not the sum of a prime and at most $3$ powers of $2$, and pointed out that parity considerations, coupled with the fact that there are many integers not the sum of a prime and $2$ powers of $2$ (see [9]) suggest that there exist infinitely many even integers which are not the sum of a prime and at most $3$ powers of $2$).

OPEN

Is every odd $n$ the sum of a squarefree number and a power of 2?

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 $p$).

This is equivalent to asking whether every $n$ not divisible by $4$ is the sum of a squarefree number and a power of two. Erdős thought that proving this with two powers of 2 is perhaps easy, and could prove that it is true (with a single power of two) for almost all $n$.

SOLVED

Is the set of odd integers not of the form $2^k+p$ the union of an infinite arithmetic progression and a set of density $0$?

OPEN - $500

If $A\subseteq \mathbb{N}$ is such that $A+A$ contains all but finitely many integers then $\limsup 1_A\ast 1_A(n)=\infty$.

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 the hypothesis $\lvert A\cap [1,N]\rvert \gg N^{1/2}$ for all large $N$ suffices.

Erdős and Sárközy conjectured the stronger version that if $A=\{a_1<a_2<\cdots\}$ and $B=\{b_1<b_2<\cdots\}$ with $a_n/b_n\to 1$ are such that $A+B=\mathbb{N}$ then $\limsup 1_A\ast 1_B(n)=\infty$.

See also [40].

SOLVED - $100

Is there an explicit construction of a set $A\subseteq \mathbb{N}$ such that $A+A=\mathbb{N}$ but $1_A\ast 1_A(n)=o(n^\epsilon)$ for every $\epsilon>0$?

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.

An explicit construction was given by Jain, Pham, Sawhney, and Zakharov [JPSZ24].

OPEN

Is there a set $A\subset\mathbb{N}$ such that
\[\lvert A\cap\{1,\ldots,N\}\rvert = o((\log N)^2)\]
and such that every large integer can be written as $p+a$ for some prime $p$ and $a\in A$? Can the bound $O(\log N)$ be achieved?

Erdős [Er54] proved that such a set $A$ exists with $\lvert A\cap\{1,\ldots,N\}\rvert\ll (\log N)^2$ (improving a previous result of Lorentz [Lo54] who achieved $\ll (\log N)^3$). Wolke [Wo96] has shown that such a bound is almost true, in that we can achieve $\ll (\log N)^{1+o(1)}$ if we only ask for almost all integers to be representable.

OPEN

Let $A\subset\mathbb{N}$ be such that every large integer can be written as $n^2+a$ for some $a\in A$ and $n\geq 0$. What is the smallest possible value of
\[\limsup \frac{\lvert A\cap\{1,\ldots,N\}\rvert}{N^{1/2}}?\]

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

SOLVED

Let $B\subseteq\mathbb{N}$ be an additive basis of order $k$ with $0\in B$. Is it true that for every $A\subseteq\mathbb{N}$ we have
\[d_s(A+B)\geq \alpha+\frac{\alpha(1-\alpha)}{k},\]
where $\alpha=d_s(A)$ and
\[d_s(A) = \inf \frac{\lvert A\cap\{1,\ldots,N\}\rvert}{N}\]
is the Schnirelmann density?

Erdős [Er36c] proved this is true with $k$ replaced by $2k$ in the denominator.

Ruzsa has observed that this follows immediately from the stronger fact proved by Plünnecke [Pl70] that (under the same assumptions) \[d_S(A+B)\geq \alpha^{1-1/k}.\]

OPEN - $500

Is there $A\subseteq \mathbb{N}$ such that
\[\lim_{n\to \infty}\frac{1_A\ast 1_A(n)}{\log n}\]
exists and is $\neq 0$?

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.

SOLVED

Is there a set $A\subset\mathbb{N}$ such that, for all large $N$,
\[\lvert A\cap\{1,\ldots,N\}\rvert \ll N/\log N\]
and such that every large integer can be written as $2^k+a$ for some $k\geq 0$ and $a\in A$?

Lorentz [Lo54] proved there is such a set with, for all large $N$,
\[\lvert A\cap\{1,\ldots,N\}\rvert \ll \frac{\log\log N}{\log N}N\]
The answer is yes, proved by Ruzsa [Ru72]. Ruzsa's construction is ingeniously simple:
\[A = \{ 5^nm : m\geq 1\textrm{ and }5^n\geq C\log m\}+\{0,1\}\]
for some large absolute constant $C>0$. That every large integer is of the form $2^k+a$ for some $a\in A$ is a consequence of the fact that $2$ is a primitive root of $5^n$ for all $n\geq 1$.

OPEN

Let $A\subset \mathbb{N}$ be an additive basis of order $2$. Must there exist $B=\{b_1<b_2<\cdots\}\subseteq A$ which is also a basis such that
\[\lim_{k\to \infty}\frac{b_k}{k^2}\]
does not exist?

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

OPEN

For $r\geq 2$ let $h(r)$ be the maximal finite $k$ such that there exists a basis $A\subseteq \mathbb{N}$ of order $r$ (so every large integer is the sum of at most $r$ integers from $A$) and exact order $k$ (i.e. $k$ is minimal such that every large integer is the sum of exactly $k$ integers from $A$). Find the value of
\[\lim_r \frac{h(r)}{r^2}.\]

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 $\geq 7$).

SOLVED

Let $A\subseteq \mathbb{N}$ be an additive basis (of any finite order) such that $\lvert A\cap \{1,\ldots,N\}\rvert=o(N)$. Is it true that
\[\lim_{N\to \infty}\frac{\lvert (A+A)\cap \{1,\ldots,N\}\rvert}{\lvert A\cap \{1,\ldots,N\}\rvert}=\infty?\]

The answer is no, and a counterexample was provided by Turjányi [Tu84]. This was generalised (to the replacement of $A+A$ by the $h$-fold sumset $hA$ for any $h\geq 2$) by Ruzsa and Turjányi [RT85]. Ruzsa and Turjányi do prove (under the same hypotheses) that
\[\lim_{N\to \infty}\frac{\lvert (A+A+A)\cap \{1,\ldots,3N\}\rvert}{\lvert A\cap \{1,\ldots,N\}\rvert}=\infty,\]
and conjecture that the same should be true with $(A+A)\cap \{1,\ldots,2N\}$ in the numerator.

OPEN

The restricted order of a basis is the least integer $t$ (if it exists) such that every large integer is the sum of at most $t$ distinct summands from $A$. What are necessary and sufficient conditions that this exists? Can it be bounded (when it exists) in terms of the order of the basis? What are necessary and sufficient conditions that this is equal to the order of the basis?

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?

OPEN

Is there a sequence $A=\{1\leq a_1<a_2<\cdots\}$ such that every integer is the sum of some finite number of consecutive elements of $A$? Can the number of representations of $n$ in this form tend to infinity with $n$?

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.