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 \[d_n> d_{n+1}>\cdots >d_{n+m}.\]
For the infinite version see [12].
The answer is no, as proved by Filaseta, Ford, Konyagin, Pomerance, and Yu [FFKPY07], who (among other results) prove that if \[1< C \leq N^{\frac{\log\log\log N}{4\log\log N}}\] then, for any $N\leq n_1<\cdots< n_k\leq CN$, the density of integers not covered for any fixed choice of residue classes is at least \[\prod_{i}(1-1/n_i)\] (and this density is achieved for some choice of residue classes as above).
An explicit construction was given by Jain, Pham, Sawhney, and Zakharov [JPSZ24].
Burr and Erdős [BuEr85] showed that there exists a constant $c>0$ such that it cannot be true that \[\lvert A\cap \{1,\ldots,N\}\rvert \leq c(\log N)^2\] for all large $N$ and that there exists a Ramsey $2$-complete $A$ such that for all large $N$ \[\lvert A\cap \{1,\ldots,N\}\rvert < (2\log_2N)^3.\] Improve either of these bounds.
Erdős was 'almost certain' that if $A$ is the set of powers of $2$ then no such $c$ exists (although he conjectured that $n$ vertices and average degree $\gg (\log n)^{C}$ suffices for some $C=O(1)$). If $A$ is the set of squares (or the set of $p\pm 1$ for $p$ prime) then he had no guess.
Solved by Verstraëte [Ve05], who gave a non-constructive proof that such a set $A$ exists.
Liu and Montgomery [LiMo20] proved that in fact this is true when $A$ is the set of powers of $2$ (more generally any set of even numbers which doesn't grow too quickly) - in particular this contradicts the previous belief of Erdős.
This problem is #4 in Ramsey Theory in the graphs problem collection.
A similar question can be asked for other even cycles.
See also [666] and the entry in the graphs problem collection.
If this fails for $4$, perhaps there is some constant for which it holds?
Erdős suggested this as an approach to solve [96]. Indeed, if this problem holds for $k+1$ vertices then, by induction, this implies an upper bound of $kn$ for [96].
The answer is no if we omit the requirement that the polygon is convex (I thank Boris Alexeev and Dustin Mixon for pointing this out), since for any $d$ there are graphs with minimum degree $d$ which can be embedded in the plane such that each edge has length one (for example one can take the $d$-dimensional hypercube graph on $2^d$ vertices). One can then connect the vertices in a cyclic order so that there are no self-intersections and no three consecutive vertices on a line, thus forming a (non-convex) polygon.
Erdős [Er94b] wrote 'I could not prove it but felt that it should not be hard. To my great surprise both B. H. Sendov and M. Simonovits doubted the truth of this conjecture.' In [Er94b] he offers \$100 for a counterexample but only \$50 for a proof.
The stated problem is false for $n=4$, for example taking the points to be vertices of a square. The behaviour of such sets for small $n$ is explored by Bezdek and Fodor [BeFo99].
See also [103].
Grünbaum [Gr76] constructed an example with $\gg n^{3/2}$ such lines. Erdős speculated this may be the correct order of magnitude. This is false: Solymosi and Stojaković [SoSt13] have constructed a set with no five on a line and at least \[n^{2-O(1/\sqrt{\log n})}\] many lines containing exactly four points.
See also [102]. A generalisation of this problem is asked in [588].
This was solved by Beck [Be91], who proved that there exists some $c>0$ such that \[\max_{n\leq N} M_n > N^c.\]
This is true if $A$ is unbounded or dense in some interval. It therefore suffices to prove this when $A=\{a_1>a_2>\cdots\}$ is a countable strictly monotone sequence which converges to $0$.
Steinhaus [St20] has proved this is false whenever $A$ is a finite set.
This conjecture is known in many special cases (but, for example, it is is open when $A=\{1,1/2,1/4,\ldots\}$. For an overview of progress we recommend a nice survey by Svetic [Sv00] on this problem.
It may be true that there are at least $n^{1-o(1)}$ many such distances. In [Er97e] Erdős offers \$100 for 'any nontrivial result'.
In [Er84] Erdős speculates that perhaps there are $\geq (1+o(1))kn/6$ many such lines, but says 'perhaps [this] is too optimistic and one should first look for a counterexample'. The constant $1/6$ would be best possible here, since there are arrangements of $n$ points with no four points on a line and $\sim n^2/6$ many lines containing three points (see Burr, Grünbaum, and Sloane [BGS74] and Füredi and Palásti [FuPa84]).
More generally, Bose and Chowla conjectured that the maximum size of $A\subseteq \{1,\ldots,N\}$ with all $r$-fold sums distinct (aside from the trivial coincidences) then \[\lvert A\rvert \sim N^{1/r}.\] This is known only for $r=2$ (see [30]).
Note that there are $\sim 2^{\binom{n}{2}}/n!$ many non-isomorphic graphs on $n$ vertices (folklore, often attributed to Pólya), and hence the bound in the problem statement is trivially best possible.
Erdős believed Brouwer's construction was essentially best possible, but Spencer suggested that $\gg \frac{2^{\binom{n}{2}}}{n!}$ may be possible. Erdős offered \$100 for a construction and \$25 for a proof that no such construction is possible.
Frankl and Rödl [FrRo86] proved $N\leq 7\times 10^{11}$, which Spencer [Sp88] improved to $\leq 3\times 10^{9}$. This resolved the initial challenge of Erdős [Er75d] to beat $10^{10}$.
Lu [Lu07] proved $N\leq 9697$ vertices. The current record is due to Dudek and Rödl [DuRo08] who proved $N\leq 941$ vertices. For further information we refer to a paper of Radziszowski and Xu [RaXu07], who prove that $N\geq 19$ and speculate that $N\leq 127$.
The restriction to $k\geq 4$ is necessary since Sylvester has shown that $f_3(n)= n^2/6+O(n)$. (See also Burr, Grünbaum, and Sloane [BGS74] and Füredi and Palásti [FuPa84] for constructions which show that $f_3(n)\geq(1/6+o(1))n^2$.)
For $k\geq 4$, Kárteszi [Ka] proved \[f_k(n)\gg_k n\log n.\] Grünbaum [Gr76] proved \[f_k(n) \gg_k n^{1+\frac{1}{k-2}}.\] Erdős speculated this may be the correct order of magnitude, but Solymosi and Stojaković [SoSt13] give a construction which shows \[f_k(n)\gg_k n^{2-O_k(1/\sqrt{\log n})}\]
Gallai was the first to consider problems of this type, and observed that $g(2)=2$ and $g(3)\geq 4$.
In [Er92c] Erdős offers '100 dollars or 1000 rupees', whichever is more, for a proof or disproof. (In 1992 1000 rupees was worth approximately \$38.60.)
Erdős and Surányi similarly asked what is the smallest $c_n\geq 1$ such that in any interval $I\subset [0,\infty)$ of length $c_n\max(A)$ there exists some $B\subseteq I\cap \mathbb{N}$ with $\lvert B\rvert=n$ such that \[\prod_{a\in A} a \mid \prod_{b\in B}b.\] They prove $c_2=1$ and $c_3=\sqrt{2}$, but have no good upper or lower bounds in general.
See also [709].
Is it true that, if $P_n$ is the path of length $n$, then \[\hat{R}(P_n)/n\to \infty\] and \[\hat{R}(P_n)/n^2 \to 0?\]
Answered by Beck [Be83b], who proved that in fact $\hat{R}(P_n)\ll n$.