Let $F\subset\mathbb{Z}$ be a sifting set composed of $n\geqslant 2$ pairs of residue classes defined as $$F=\bigcup\nolimits ^{n}_{i=1}\{x:x\equiv \pm m_{i} \pmod{p_{i}}\}$$ where $p_{i}$ are distinct odd primes verifying $5\leqslant p_{1} < ...< p_{n}$, and with $m_{i}=\left\lfloor \frac{p_{i} +1}{6}\right\rfloor $ for all $i$. The complement of $F$ is denoted $F^{c} =\mathbb{Z} \backslash F$.
Let $A\subset\mathbb{Z}$ be a set of $N$ consecutive integers, i.e. $A=\{M+1,...,M+N\}$ for some integer $M$, show that for sufficiently large $n$ $$N\geqslant 2p_{n} \Longrightarrow A\cap F^{c} \neq \varnothing $$
Let's take an example to get an intuition for the problem. Say we work with $p_1=5$ and $p_2=7$. The pattern created by $F$ is periodic, so we only need to look at $\{0,34\}$, and we mark in red the cells corresponding to $\pm 1 \pmod{5}$ and $\pm 1 \pmod{7}$ $$\unicode{11035} \color{red}{\unicode{11035}} \unicode{11035} \unicode{11035} \color{red}{\unicode{11035}} \unicode{11035} \color{red}{\unicode{11035}} \unicode{11035} \color{red}{\unicode{11035}} \color{red}{\unicode{11035}} \unicode{11035} \color{red}{\unicode{11035}} \unicode{11035} \color{red}{\unicode{11035}} \color{red}{\unicode{11035}} \color{red}{\unicode{11035}} \color{red}{\unicode{11035}} \unicode{11035} \unicode{11035} \color{red}{\unicode{11035}} \color{red}{\unicode{11035}} \color{red}{\unicode{11035}} \color{red}{\unicode{11035}} \unicode{11035} \color{red}{\unicode{11035}} \unicode{11035} \color{red}{\unicode{11035}} \color{red}{\unicode{11035}} \unicode{11035} \color{red}{\unicode{11035}} \unicode{11035} \color{red}{\unicode{11035}} \unicode{11035} \unicode{11035} \color{red}{\unicode{11035}} $$
Is there a stretch of $2\cdot 7 = 14$ cells (wrapping around if necessary) with no black ones?
The next level is to play with $\{5,7,11\}$. The periodic pattern of $F$ will have length $385$, and there will be $105$ black cells distributed in it, after we have marked red the cells corresponding to $\pm 1 \pmod{5}$, $\pm 1 \pmod{7}$ and $\pm 2 \pmod{11}$. And again, no stretches of length $22$ will be "black-cell-less".
A few notes
- W.l.o.g. the $\{p_i\}$ set may be assumed to be the factors of $\frac{q_{n+2}\#}{6}$ (where $q_k$ is the $k$-th prime in $\mathbb{N}$)
- A simple reasoning offers the expectation that the cardinality of $A\cap F^{c}$ should be on average $N\prod\nolimits ^{n}_{k=1}\left( 1-\frac{2}{p_{k}}\right)$, and since $2p_{n}\prod\nolimits ^{n}_{k=1}\left( 1-\frac{2}{p_{k}}\right)$ grows like $\Theta \left(\frac{x}{\log^{2} x}\right)$, it would seem that the claim should also be true in a more general form with arbitrary pairs of residue classes. The specific ones mentioned above are what I'm actually interested in, and may perhaps offer a shortcut to a proof for that specific case (the hope is that their somewhat regular spread in their respective moduli confers a low-ish discrepancy to the distributions). Note: the product expression has been previously studied on MSE in an apparently different context, but as can be seen below in the "backstory", this product appearing here isn't surprising after all.
On the topic of due diligence, I have attempted to solve the problem with combinatorial means (again, that's why the residue classes are specified). I guess I hoped I would be able to find some brilliant probabilistic argument. So far, my attempts at impersonating Paul Erdös have been met with perfect utter failure (which is perfectly unsurprising as well). I also looked into uncertainty principles for DFTs with similar (non-)results, both with Hirschmann entropic uncertainty and with deterministic uncertainty (see this quite unpopular question of mine for an example of my unsuccessful attempts in that direction). I briefly considered looking at the problem from the angle of hypergraphs, but most theorems are related to discrepancies over all possible colorings, and not to discrepancies over a specific one.
I'm left with the obvious approach: sieve theory. Here too, I diligently checked the site for previously asked questions and actually found a related one. The statement of mkultra's question is a generalisation of the one above. I even upvoted reuns's answer on account that it is technically correct: the provided lower bound can be non-trivial as requested. Unfortunately, I believe this is Legendre's sieve, and this is completely useless for achieving my goal.
Finally, I tried to educate myself, of course. But I have to admit, the technicalities are melting my brain. I believe I understand that applicable sieves would be either Rosser's or some form of Buchstab-iterated Selberg ones. But I can't seem to be able to figure out how to apply them. Just to give an example: is Theorem 11.13 in Richert's Lectures on Sieve Methods applicable and/or useful here, and if so, how?
Background on the problem
[I have checked on meta, and the consensus seems to be that I should talk about this right here, rather than creating a separate self-answered question. And to follow another advice I found on meta as well, I will keep it as short as I can and do a lot of claiming without proving. If something in the following warrants further study or explanations, feel free to ask a specific question about it on Math.SE]
I arrived at the above problem statement while studying the following cardinality
$$ \left\vert\mathbb{N} \backslash \bigcup\nolimits_{k\in\mathbb{N}} S_k \right\vert \tag{1} $$
where $S_k$ is a union of arithmetic progressions defined as
$$S_{k}=\left(\begin{aligned} & \{(5k-1) +( 6k-1) x\}^{\infty }_{x=0}\\ \cup & \{(7k-1)+(6k-1)x\}^{\infty}_{x=0}\\ \cup & \{(5k+1)+(6k+1)x\}^{\infty}_{x=0}\\ \cup & \{(7k+1)+(6k+1)x\}^{\infty}_{x=0} \end{aligned}\right) \tag{2}$$
Visually, grouped by moduli and stacked above/under each other, the $S_k$ arithmetic progressions look like this:
$S_16$ up to 100" />
Notice that some columns are empty. The question of whether the cardinality $(1)$ is finite or not simply translates to: are there infinitely many of those empty columns, or not?
A first observation (and proving it is trivial): the composite moduli are redundant. For example:
All columns hit by residues $\bmod{35}$ already have hits in $\bmod{5}$ and $\bmod{7}$.
Another pattern is a bit more difficult to discern, but not much harder to prove:
Therefore the spots marked in black are the only ones we need to care about.
As can be shown, columns $z$ in the range $6m^2-2m \leqslant z < 6(m+1)^2-2(m+1)$ can only be affected by moduli up to $6m+1$. The ranges span over $12m+4 = 2(6m+1)+2$ columns. In other words, we're interested in a non-trivial lower bound for a sieve with pairs of residue classes modulo primes from $5$ up to (let's re-label) $p_n$, and we're applying that sieve on a range consecutive integers of size $2p_n$ (plus tax).
Therefore it would follow from the unproven claim I'm asking about, that the cardinality $(1)$ is in fact infinite.
Background on the background
Some observant readers may have noticed that if they take the indices of the empty columns in the screenshots above and feed them to the OEIS search box, the returned hit is a specific sequence, which is easily explained.
The set $\mathbb{N} \backslash \bigcup\nolimits_{k\in\mathbb{N}} S_k$ mentioned in $(1)$ is the set of all $z \in \mathbb{N}$ that cannot be represented as one of the following forms: $$\begin{aligned} 6xy+5x+5y+4\\ 6xy+7x+5y+6\\ 6xy+7x+7y+8\\ \end{aligned} \tag{3} $$ for all $(x,y) \in (\mathbb{N} \cup \{0\})^2$. Equivalently (just a matter of shifting variables by $1$), this can be stated as $\mathbb{N} \backslash \bigcup\nolimits_{k\in\mathbb{N}} S_k$ being the set of all $z \in \mathbb{N}$ that cannot be represented as: $$6xy \pm x \pm y \tag{4}$$ for all $(x,y) \in (\mathbb{Z} \backslash \{0\})^2$. In turn (this can also be shown by just a change of coordinates), it is the set of all $z \in \mathbb{N}$ such that $$ X^2-Y^2 = 6z \pm 1 \tag{5}$$ both have only one solution in $\mathbb{N}^2$. Note: many other ways to derive the same fact starting from $(4)$ have been discussed in the past on MSE, like here and here.
So... yeah, this is an attack on that good ol' conjecture. Clumsy maybe? Probably foolish and futile, I'm sure. Nevertheless, I find this approach interesting, because it gives me a glimmer of hope that the problem could be solved with purely combinatorial or probabilistic means, perhaps even "just" harmonic analysis. There's also this question of whether modern sieve theoretic tools, applied under this slightly different angle, wouldn't in fact already be able to crack this tough nut. Granted, it could be that the proposed setup above is in fact equivalent to a direct sieve on the twin primes, but my intuition tells me it's not.
There is an alternative representation of your patterns (when turning by 90 degrees):
Twin primes in special divisor plane
Note: In the special divisor plane for twin primes, each divisor point describes a solution to the divisor relation $x|(6y\pm1)$, i.e. $x$ divides $(6y+1)$ OR $(6y-1)$.
As well as in the general divisor plane the points are connected by parabolas !
Parabolas in the special divisor plane for twin primes
Each set of 6 parabolas is written as (with $i \ge 1$):
$\frac{-x² - 1}{6} + \frac{6i - 1}{3x}$
$\frac{-x² - 1}{6} + \frac{6i}{3x}$
$\frac{-x² - 1}{6} + \frac{6i + 1}{3x}$
$\frac{-x² - 1}{6} + \frac{6i + 2}{3x}$
$\frac{-x² - 1}{6} + \frac{6i + 3}{3x}$
$\frac{-x² - 1}{6} + \frac{6i + 4}{3x}$
There could be an approach to a combinatorial sieve for twin primes. Above all, the connecting parables look promising.