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In this post we denote the sequence of prime numbers as $p_k$ for integers $k\geq 1$. I don't know if the following definition is in the literature.

Definition. We define the $\theta$-strong primes, or strong primes at level $\theta$, as the sequence of those prime numbers $p_n$ that satisfy the inequality $$p_n>\theta\, p_{n-1}+(1-\theta)p_{n+1}\tag{1}$$ for some fixed real number $0<\theta<1$.

Remark. My definition arises as a generalization of the known as strong primes. See the definition in number theory of strong primes from the Wikipedia Strong prime.

Question. I would like to know, if my question is interesting, if it is possible to prove that there exists some $\hat\theta$ for which the corresponding sequence of prime numbers has finitely/infinitely many terms. Many thanks.

Here I emphasize that $0<\theta<1$.

I have no intuition if my question is interesting, I don't know what can be a more interesting question at research level about the study of these sequences $(1)$ and their corresponding $\theta$. I did some calculations for $\theta\neq\frac{1}{2}$ a rational number.

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  • $\begingroup$ Please if the question does not have the best mathematical content feel free to comment it. After I wrote the definition I thought it might be interesting to ask about this definition. I don't know what can be a more interesting question at research level about the study of these sequences related to their corresponding $\theta$. $\endgroup$ – user142929 Jul 1 '20 at 20:58
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    $\begingroup$ since we know there are infinitely many consecutive primes with finite gap $C$ (I think smallest $C$ is $246$ now but not sure), it is trivial that if $2\theta > (1-\theta)C$ the $p_n$'s for which $p_{n+1}-p_n \le C$ work since $p_n-p_{n-1} \ge 2$ so $\theta(p_n-p_{n-1}) >C(1-\theta) \ge (1-\theta)(p_{n+1}-p_n)$; for $246$ we get $1 > \theta > 123/124$ satisfies the requirement with infinite $p_n$'s $\endgroup$ – Conrad Jul 2 '20 at 0:58
  • $\begingroup$ Many thanks for your excellent contribution, feel free to made your comment as an answer @Conrad $\endgroup$ – user142929 Jul 2 '20 at 5:39
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Regarding any $\hat\theta$ for which the prime numbers sequence has finitely/infinitely many terms, consider one which has only finitely many terms. There would then exist a prime index $m$ for which all $n \gt m$ gives

$$p_n \le \hat\theta\, p_{n-1} + (1 - \hat\theta)p_{n+1} \tag{1}$$

Using the standard definition of prime gaps of

$$g_n = p_{n+1} - p_{n} \tag{2}\label{eq2A}$$

we then have

$$\begin{equation}\begin{aligned} p_n & \le \hat\theta(p_{n} - g_{n-1}) + (1 - \hat\theta)(p_{n} + g_{n}) \\ p_n & \le \hat\theta p_{n} - \hat\theta g_{n-1} + (1 - \hat\theta)p_{n} + (1 - \hat\theta)g_{n} \\ \hat\theta g_{n-1} & \le (1 - \hat\theta)g_{n} \\ g_n & \ge \left(\frac{\hat\theta}{1 - \hat\theta}\right)g_{n-1} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

If $\hat\theta \gt 0.5$, then $\frac{\hat\theta}{1 - \hat\theta} \gt 1$, so \eqref{eq3A} shows the prime gaps are strictly increasing for all $n \gt m$. However, this contradicts that there are infinitely many prime gaps of at most $246$ (e.g., see Bounded gaps between primes). If $\hat\theta = 0.5$ instead, then $\frac{\hat\theta}{1 - \hat\theta} = 1$, so the prime gaps are non-decreasing, but this is also not possible since prime gaps can become arbitrarily large and, thus, must decrease later to become the smaller prime gaps, e.g., those at most $246$.

This means the original assumption of there being only a finite number of primes in the sequence must be incorrect, i.e., there are infinitely many $\theta$-strong primes for any $\hat\theta \ge 0.5$.

Update: The PDF version of the arXiv article On the ratio of consecutive gaps between primes, near the bottom of page $8$, states about Erdős

He mentioned 60 years ago [Erd5]: "One would of course conjecture that

$$\underset{n \to \infty}{\lim \inf}\frac{d_{n+1}}{d_n} = 0 \; \text{ and } \; \underset{n \to \infty}{\lim \sup}\frac{d_{n+1}}{d_n} = \infty \tag{2.2}$$

but these conjectures seem very difficult to prove." Based on a generalization of the method of Zhang [Zha] the author proved (2.2) in [Pin2]

where [Pin2] is

J. Pintz, Polignac numbers, conjectures of Erdős on gaps between primes and the bounded gap conjecture. arXiv: 1305.6289 [math.NT] 27 May 2013.

i.e., here.

Note their $d_n$ is the same as $g_n$ in \eqref{eq2A}. If the first part of (2.2) is true, then no matter how close $\frac{\hat\theta}{1 - \hat\theta}$ is to $0$, \eqref{eq3A} cannot always being true for all $n \gt m$ for any integer $m$. This means there are infinitely many $\theta$-strong primes for all $0 \lt \hat\theta \lt 1$.

I haven't read the article or references to try to verify the validity of the claim. However, note the article seems to be basically very similar to at least part of the Springer Link book On the Ratio of Consecutive Gaps Between Primes, but I haven't paid to get a chapter or the entire e-book, or a physical copy of the book, to check on this.

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  • $\begingroup$ Many thanks for your excellent answer, in a similar way than previous answer I'm understanding all details. Many thanks, are very good answers. $\endgroup$ – user142929 Jul 2 '20 at 16:48
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    $\begingroup$ @user142929 You're welcome. Note I've updated my answer with a reference to an article which contains a result that shows all $\theta$-strong prime sequences have an infinite # of primes. $\endgroup$ – John Omielan Jul 2 '20 at 19:42
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As requested I put my comment as a (partial) answer:

Since we know there are infinitely many consecutive primes with finite gap $C$ (smallest proved $C$ as of now seems to be $246$), any $\theta$ for which $2\theta >(1-\theta)C$ works for the $p_n$'s precisely the primes for which $p_{n+1}-p_n \le C, n \ge 2$ since then $p_n-p_{n-1} \ge 2$ so:

$\theta(p_n-p_{n-1}) \ge 2\theta > C(1-\theta) \ge (1-\theta)(p_{n+1}-p_n)$

For $C=246$ we get that $123/124 < \theta <1$ satisfies the requirement of the OP with infinitely many $p_n$'s

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  • $\begingroup$ Many thanks for your excellent answer. I'm going to study the details. $\endgroup$ – user142929 Jul 2 '20 at 16:44

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