See also [687].
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 \[d_n> d_{n+1}>\cdots >d_{n+m}.\]
The sequence of such numbers is A006286 in the OEIS.
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 $2$).
Tao [Ta23] has proved that this series does converge assuming a strong form of the Hardy-Littlewood prime tuples conjecture.
In [Er98] Erdős further conjectures that \[\sum_{n=1}^\infty (-1)^n \frac{1}{n(p_{n+1}-p_n)}\] converges and \[\sum_{n=1}^\infty (-1)^n \frac{1}{p_{n+1}-p_n}\] diverges. He further conjectures that \[\sum_{n=1}^\infty (-1)^n \frac{1}{n(p_{n+1}-p_n)(\log\log n)^c}\] converges for every $c>0$, and reports that he and Nathanson can prove this for $c>2$ (and conditionally for $c=2$).
Blecksmith, Erdős, and Selfridge [BES99] proved that the number of such primes is \[\ll_A \frac{x}{(\log x)^A}\] for every $A>0$, and Elsholtz [El03] improved this to \[\ll x\exp(-c(\log\log x)^2)\] for every $c<1/8$.
Solved by Tao [Ta23b], who proved that \[ \lvert A\rvert \leq \left(1+O\left(\frac{(\log\log x)^5}{\log x}\right)\right)\pi(x).\]
In [Er95c] Erdős further asks about the situation when $\phi(a_1)\leq \cdots \leq \phi(a_t)$.
The existence of such progressions for small $k$ has been verified for $k\leq 10$, see the Wikipedia page. It is open, even for $k=3$, whether there are infinitely many such progressions.
See also [219].
The values of the sum are listed at A074741 on the OEIS.
The sequence of values of $f(n)$ is A109925 on the OEIS.
See also [237].
The limit is infinite for a finite set of primes, which follows from a theorem of Pólya [Po18], that if $P(n)$ is a quadratic integer polynomial without repeated roots then as $n\to \infty$ the largest prime factor of $f(n)$ also approaches infinity. Indeed, if $P$ is a finite set of primes and $(a_i)$ is the set of integers divisible only by primes in $P$, and $a_{i+1}-a_i$ is bounded, then there exists some $k$ such that $a_{i+1}=a_i+k$ infinitely often, which contradicts Pólya's theorem with $f(n)=n(n+k)$.
Tijdeman [Ti73] proved that, if $P$ is a finite set of primes, then \[a_{i+1}-a_i \gg \frac{a_i}{(\log a_i)^C}\] for some constant $C>0$ depending on $P$.
Tijdeman [Ti73] resolved this question, proving that, for any $\epsilon>0$, there exists an infinite set of primes $P$ such that, with $a_i$ defined as above, \[a_{i+1}-a_i \gg a_i^{1-\epsilon}.\]
See also [368].
For $k=1$ or $k=2$ any set $A$ such that $\sum_{n\in A}\frac{1}{n}=\infty$ has this property.
Indeed, we apply this with $k=q=d$ and $a=1$ and let $p_m,\ldots,p_{m+{d-1}}$ be consecutive primes all congruent to $1$ modulo $d$, with $m>n+1$. If $p_{n+1}+\cdots+p_{m-1}\equiv r\pmod{d}$ with $1\leq r\leq d$ then \[d \mid p_{n+1}+\cdots +p_m+\cdots+p_{m+d+r-1}.\]
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].
Can one prove this is false if we replace $k^2+1$ by $e^{(1+\epsilon)\sqrt{k}}+C_\epsilon$, for all $\epsilon>0$, where $C_\epsilon>0$ is some constant?
Erdős observed that Cramer's conjecture \[\limsup_{k\to \infty} \frac{p_{k+1}-p_k}{(\log k)^2}=1\] implies that for all $\epsilon>0$ and all sufficiently large $n$ there exists some $k$ such that \[p(n+k)>e^{(1-\epsilon)\sqrt{k}}.\] There is now evidence, however, that Cramer's conjecture is false; a more refined heuristic by Granville [Gr95] suggests this $\limsup$ is $2e^{-\gamma}\approx 1.119\cdots$, and so perhaps the $1+\epsilon$ in the second question should be replaced by $2e^{-\gamma}+\epsilon$.
Let $f(n)$ be the smallest $k$ such that $u>n^2$. Give bounds for $f(n)$.
Mahler's theorem implies $f(n)\to \infty$ as $n\to \infty$, but is ineffective, and so gives no bounds on the growth of $f(n)$.
One can similarly ask for estimates on the smallest integer $f(n,k)$ such that if $m$ is the factor of $\binom{n}{k}$ containing all primes $\leq f(n,k)$ then $m > n^2$.
See also [677].
Is it true that $r(x)\to \infty$? Or even $r(x)/\log x \to \infty$?
This is probably false, since Hensley and Richards [HeRi73] have shown that this is false assuming the Hardy-Littlewood prime tuples conjecture.
Erdős [Er85c] reports Straus as remarking that the 'correct way' of stating this conjecture would have been \[\pi(x+y) \leq \pi(x)+2\pi(y/2).\] Clark and Jarvis [ClJa01] have shown this is also incompatible with the prime tuples conjecture.
In [Er85c] Erdős conjectures the weaker result (which in particular follows from the conjecture of Straus) that \[\pi(x+y) \leq \pi(x)+\pi(y)+O\left(\frac{y}{(\log y)^2}\right),\] which the Hensley and Richards result shows (conditionally) would be best possible. Richards conjectured that this is false.
Erdős and Richards further conjectured that the original inequality is true almost always - that is, the set of $x$ such that $\pi(x+y)\leq \pi(x)+\pi(y)$ for all $y<x$ has density $1$. They could only prove that this set has positive lower density.
They also conjectured that for every $x$ the inequality $\pi(x+y)\leq \pi(x)+\pi(y)$ is true provided $y \gg (\log x)^C$ for some large constant $C>0$.
Hardy and Littlewood proved \[\pi(x+y) \leq \pi(x)+O(\pi(y)).\] The best known in this direction is a result of Montgomery and Vaughan [MoVa73], which shows \[\pi(x+y) \leq \pi(x)+2\frac{y}{\log y}.\]
Estimate $h(n)$.
It is a classical fact that \[\limsup_{n\to \infty}\omega(n)\frac{\log\log n}{\log n}=1.\]