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Let $n\geq 4$ be some integer, and let $k$ be largest prime index such that $p_k \leq \sqrt{2n}$.

We assert there is some integer $x$ in $\frac{n}{2} < x < n$ such that we have a set of congruences for primes from $p_2$ to $p_k$:

$$ \begin{cases} x &\not\equiv 0 \pmod{3}\\ x+1 &\not\equiv 0 \pmod{5}\\ x+2 &\not\equiv 0 \pmod{7}\\ x+4 &\not\equiv 0 \pmod{11}\\ x+5 &\not\equiv 0 \pmod{13}\\ \vdots\\ x+\frac{p_k-3}{2} &\not\equiv 0 \pmod{p_k} \end{cases} $$

Any valid $x$ implies $2x-3\in\mathbb P$.

I know essentially nothing of combinatorics, so I'm certain the following reasoning is off, but I need somebody to explain where I've gone wrong. The motivation behind this is seeking an alternative proof for Bertrand's Postulate. Informally, I'm trying to argue that in every $(n/2,n)$, you'll be able to find something like $x=7\perp 3$ and $x+1=8 \perp 5$ (where $\perp$ is coprime), which is sufficient to give you $5+6=11\in\mathbb P$. The larger the $n$, the more prime factors need to be checked for coprimality, up to a maximum of $\sqrt{2n}$.

I'd like to be able to construct an expression showing a worst-case scenario, and ideally that some $x$ will exist even in that case. One approach is

$$s=\frac{n}{2} \prod_{i=2}^{k}{\frac{p_i-1}{p_i}}$$

where $s$ would be the expected number of valid $x$ values in $(\frac{n}{2},n)$, although to be clear, I am looking for a guaranteed minimum rather than an average expectation. To that end, even

$$s=\frac{n}{2} \prod_{i=2}^{k}{\frac{p_i-2}{p_i}}$$

seems to yield a positive result, albeit by a slim margin.

I also tried starting with $s:=n/2$ and iterating

$$s \leftarrow \left\lfloor s\cdot \frac{p_i-1}{p_i} \right\rfloor -1$$

or similarly,

$$s \leftarrow s-\left\lceil \frac{s}{p_i} \right\rceil$$

for $2\leq i \leq k$, again attempting to err on the side of worst-case. All of these approaches give me the approximate result I want (a steadily growing positive number), but since it's unrealistic for the result to be this easy, I know I must be oversimplifying things somehow...

Trevor
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    Bertrand (or one of the stronger versions of Bertrand) lets you find a prime $\psi$ slightly greater than $n$, and then $x=\frac {\psi+3}2$ should fit your requirements, no? – lulu Feb 11 '20 at 23:32
  • @lulu Alas, Bertrand is the thing I'm trying to find an alternate approach to with this. – Trevor Feb 11 '20 at 23:40
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    I think you should make that clear in your post. – lulu Feb 11 '20 at 23:42
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    I take it you're looking at $\prod_i((p_i-1)/p_i)$ numerically and getting encouraging results. But rigorous results on this product are harder to get, maybe just as hard as proving Bertrand some other way. – Gerry Myerson Feb 12 '20 at 00:46
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    But you've been here before: https://math.stackexchange.com/questions/3474958/a-trivial-proof-of-bertrands-postulate where your "trivial proof" went on for pages and pages and never did settle down. What's wrong with the usual approaches to Bertrand? – Gerry Myerson Feb 12 '20 at 00:48
  • I'm a dog with a bone, what can I say? – Trevor Feb 12 '20 at 01:23
  • Actually, I did try to address that here: https://math.stackexchange.com/questions/3529914/soft-intuition-in-number-theory-bertrands-postulate-as-good-as-it-gets – Trevor Feb 12 '20 at 01:50

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