1 solved out of 22 shown

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

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 $1-\epsilon$.

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.

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 $0$?

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

$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 $\lvert A\cap [1,N]\rvert \gg N^{1/2}$ 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 $\limsup 1_A\ast 1_B(n)=\infty$.

$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.

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 $a\in A$?

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

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 $>1$.

Let $A\subseteq\mathbb{N}$ be an additive basis of order $k$. Is it true that for every $B\subseteq\mathbb{N}$ 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 proved this is true with $k$ replaced by $2k$ in the denominator.

$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.

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

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].

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 $\geq 7$).

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?

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.