Question in the title.

It intuitively seems absurd that $p_N - p_{N-1} \gt p_{N-1} - 3 = $ the largest gap formable from all $p_i = $ odd primes $3, \dots, p_{N-1}$.

Was wondering how difficult the proof is.

$2 p_i$ is the smallest composite divisible by $p_i$. And $p_N + 3$ is certainly a composite. Not sure if that helps : >

Thanks @PaoloLeonetti in the comments. According to the article:

In 1998, Pierre Dusart improved the result in his doctoral thesis, showing that for $k \geq 463, p_{k+1} \leq (1 + 1/(\ln^2 p_k))p_k$, ...

So we want to show that $p_{k+1} \leq 2 p_{k} - 3$ for sufficiently large $k$.

$2 p_k - 3 = (2 - \dfrac{3}{p_k})p_k$ and

$$ 2 - \dfrac{3}{p_k} \geq 1 + 1 / (\ln^2 p_k) \iff (1 - \dfrac{3}{p_k})\ln^2 p_k \geq 1 \iff \\ \ln^2 p_k \geq \dfrac{p_k}{p_k - 3} $$

the last operation being valid since $k \geq 463$ and so $3$ is much less than $p_k$ and so $1 - \dfrac{3}{p_k} \gt 0$.

Now take the exponential:

$$ \iff 8 \approx \ln(p_k) \geq e^{\frac{p_k}{p_k - 3}} \approx 2 $$ Where the approximation is at least valid enough for $k \geq 463$.

  • $\begingroup$ Have you tried anything yourself? Any known prime gaps you are familiar with? $\endgroup$ – Sean Nemetz Mar 4 '18 at 1:22
  • $\begingroup$ @SeanNemetz That's the trouble, I don't know those analytic bounds. But can we do this without those, since the gap is so absurdly huge? $\endgroup$ – OlympicComputerChairSitter Mar 4 '18 at 1:23
  • $\begingroup$ @SeanNemetz I have tried some things on paper, but they didn't lead any where. $\endgroup$ – OlympicComputerChairSitter Mar 4 '18 at 1:24
  • 3
    $\begingroup$ en.wikipedia.org/wiki/Bertrand%27s_postulate : this is nearly sufficient to your question. There is a variant which states that, for all $\varepsilon>0$, then there is a prime $p$ such that $n<p<n(1+\varepsilon)$ whenever $n$ is sufficiently large. $\endgroup$ – Paolo Leonetti Mar 4 '18 at 1:25
  • $\begingroup$ @PaoloLeonetti how do we translate that over to this? In particular Jitsuro Nagara's $n \lt p \lt \dfrac{6}{5} n$ postulate. $\endgroup$ – OlympicComputerChairSitter Mar 4 '18 at 1:36

It is known that if $n\ge 25$, there is always a prime $p$ such that $n\le p\le 6n/5$ (Jitsuro Nagura, 1952). On the other hand, your claim is $$ p_{k+1} \le 2p_k-3 $$ for all $k$. Thanks to the above result, if $p_k \ge 29$, there exists a prime $p^\star$ in the interval $[p_k+1, \frac{6}{5}(p_k+1)]$. Moreover, $p^\star$ will be at least $p_{k+1}$, so $$ p_k < p_{k+1} \le p^\star \le \frac{6}{5}(p_k+1) $$ and it will be sufficient to prove that $\frac{6}{5}(p_k+1) \le 2p_k-3$, i.e., $p_k \ge \frac{21}{4}$. And this is verified if $p_k \ge 29$. For the remaining cases, check it manually.


Simply applying Bertrand's postulate in its restrictive form

For any integer ${\displaystyle n>3}$, there always exists at least one prime number ${\displaystyle p}$ with $${\displaystyle n<p<2n-2}$$

If we consider $n=p_k$ then $p_k < p_{k+1} < 2p_k-2$, but $2p_k-2$ is even so can't be a prime, thus $$p_k < p_{k+1} \leq 2p_k-3 \Rightarrow p_{k+1} +3 \leq 2p_k$$


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