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As I recall, the most common difference between consecutive primes starts with $2$, moves on to $6$, then $30$, and is conjectured to progress like the primorials over a long time, without bound.

Is there any analogous prediction about $x^2+1$ or other polynomial prime gaps?

Rushabh Mehta
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Trevor
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    IIRC it is unknown whether there are infinitely many primes of the form $x^2+1$. So kinda difficult to make a prediction like this. – Jyrki Lahtonen Dec 06 '19 at 18:58
  • polynomoal remander theorem not enough for you. –  Dec 06 '19 at 21:54
  • simply using the first two terms, and some modular arithmetic, I can tell you any $x$ that could create primes all end in 0,4, or 6. for example. –  Dec 06 '19 at 23:04

1 Answers1

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Those things are conjectured by the random model for the primes, the model on which every conjecture about primes is based. This model is saying that when $n$ is picked uniformly in $[1,N]$, for $p\le N^r$ then $n\bmod p$ is uniformly distributed in $1,\ldots, p$ and it is independent from one $p$ to the other. Of course this model is wrong for a fixed $N$, what we assume is that it gets less wrong as $N\to \infty$, ultimately giving correct predictions.

  • What is $r$ ? It is the constant $r\in [1/2,1)$ such that the probability that $n$ (picked uniformly in $[1,N]$) is prime is $$\prod_{p \le N^r} Pr(p\nmid n)=\prod_{p \le N^r} (1-p^{-1})$$

    This $r$ can be evaluated from Mertens' theorem and the PNT : it is $r=e^{-\gamma}$, the only constant such that $$\lim_{N \to \infty}\frac{\prod_{p \le N^r} (1-p^{-1})}{\pi(N)/N}=\lim_{N \to \infty}\log N\prod_{p \le N^r} (1-p^{-1})=1$$

  • The probability that $n,n+2$ are both primes is $$\prod_{p \le N^r} Pr(p\nmid n,p\nmid n+2)=(1-2^{-1})\prod_{3\le p \le N^r} (1-2 p^{-1})$$

    The conjectured number of twin primes $\le N$ is $$C_2\frac{N}{\log^2 N}\sim N(1-2^{-1})\prod_{3\le p \le N^r} (1-2 p^{-1})$$ where $C_2$ is the twin prime constant $$C_2=\lim_{N\to \infty}(\log^2 N) (1-2^{-1})\prod_{3\le p \le N^r} (1-2 p^{-1})=\lim_{N\to \infty} \frac{(1-2^{-1})\prod_{3\le p \le N^r} (1-2 p^{-1})}{\prod_{ p \le N^r} (1- p^{-1})^2}$$

  • The probability that $n^2+1$ is prime is $$\prod_{p\le N^{2r}} Pr(p\nmid n^2+1)= (1-2^{-1})\prod_{p\le N^{2r},p\equiv 1\bmod 4} (1-2 p^{-1})$$

    Thus the conjectured number of primes $n^2+1$ with $n\le N$ is $$A \frac{N}{\log N}$$ where the obtained constant is $$A = \lim_{N\to \infty}(\log N) (1-2^{-1})\prod_{p\le N^{2r},p\equiv 1\bmod 4} (1-2 p^{-1})$$ $$= \lim_{N\to \infty}\frac{(1-2^{-1})\prod_{p\le N^{2r},p\equiv 1\bmod 4} (1-2 p^{-1})}{\frac12 \prod_{p\le N^{2r}}(1-p^{-1})}$$

    And the conjectured mean distance between two consecutive such primes $n^2+1,n_2^2+1$ is $$A\log n$$

reuns
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  • all $x>2$ need to end in 0,4 or 6 to produce primes. –  Dec 07 '19 at 13:48
  • so what ? ${}{}$ – reuns Dec 08 '19 at 08:31
  • I have seen Mertens third theorem written like $\lim_{N\to\infty}\ln N\prod_{p\le N}\left(1-\frac1p\right)=e^{-\gamma}$ from there I see that writing $N^{\gamma}$ in the index instead should make the product bigger but I can't see why it should be exactly 1, could you give me some hint?, thanks – Dabed Dec 09 '19 at 20:56
  • I messed up it is $r=e^{-\gamma}$ not $r = \gamma$ (indeed they are almost equal) and it disappears in the obtained infinite products for the constants $C_2,A$ – reuns Dec 09 '19 at 21:01
  • I didn't even realize there was an error but is nice you found it what I feel confused about is why is possible to pass from $\lim_{N\to\infty}\ln N\prod_{p\le N}\left(1-\frac1p\right)=c$ to $\lim_{N\to\infty}\ln N\prod_{p\le N^c}\left(1-\frac1p\right)=1$, as far as I can understand the rest seems fine to me so not sure why you say it disappears in C_2 and A, I'm sorry and thanks again – Dabed Dec 09 '19 at 21:24
  • It is just $\lim_{N\to\infty}\ln N\prod_{p\le N}\left(1-\frac1p\right)=c \implies $ $\lim_{N\to\infty} \ln N\prod_{p\le N^c}\left(1-\frac1p\right)=\lim_{N\to\infty} \frac1c\ln N^c\prod_{p\le N^c}\left(1-\frac1p\right)$ $=\lim_{M\to\infty} \frac1c\ln M\prod_{p\le M}\left(1-\frac1p\right)=\frac{c}{c}$ – reuns Dec 09 '19 at 21:32
  • Thanks, I realize now what I was asking was pretty basic, sorry about that – Dabed Dec 09 '19 at 21:40