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A few months ago I asked this question on Mathematics Stack Exchange but it has received little attention. Perhaps the question is more applicable here.

Let $p_k$ denote the $k$th prime such that $p_1=2$, and consider the following array of coordinates: \begin{array}{c|c}x_i&2&5&11&17&23&31&\cdots\\\hline y_i&3&7&13&19&29&37&\cdots\end{array} where $i=1,2,\cdots$. Then $x_i=p_{2i-1}$ and $y_i=p_{2k}$, so we are using the first $2k$ primes.

If $y_i=\alpha+\beta x_i$ is the least squares regression line for these prime coordinates, does $\alpha$ converge as $i\to\infty$ and if so, to what value?

Note that $\beta=1+\epsilon\to1^+$ as $i\to\infty$ for some $\epsilon>0$ as $y_i>x_i$. The following table gives the value of $\alpha$ for $i=10^j$. \begin{array}{c|c}j&1&2&3&4&5&6&7&8\\\hline\alpha&0.33&2.41&4.08&6.57&8.91&11.26&13.57&15.84\end{array} It may however be too early to tell whether $\alpha$ converges as $j\le8$.

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    $\begingroup$ Do you mean that for each $k$, we use the first $k$ data points (that is, the first $2k$ primes) to determine fit coordinates $\alpha$ and $\beta$ that depend on $k$? If so, you should clarify that. $\endgroup$ Mar 31, 2019 at 18:28
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    $\begingroup$ Shouldn’t it grow like a constant times log i? $\endgroup$
    – alpoge
    Mar 31, 2019 at 20:09

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This is too long for a comment.

There is something that I do not understand (more than surely, I am missing a point) : if I perform these linear regressions, I do not obtain the results given in the post.

Forcing $\beta=1$, what I obtain for the successive values of $j=10^i$ is $$\begin{array}{c|c}&j&1&2&3&4&5&6&7&\\\hline&\alpha&3.30&5.97&8.81&11.36&13.80&16.23&18.68&\end{array}$$ which looks like a quadratic in $\log(j)$.

Just to clarify, ig give below the syntax I used for genarating the data for Mathematica

  `data[j_]:=Table[{Prime[2k-1],Prime[2k]},{k,1,10^j}]`

Edit

Doing the same with $5^j$ instead of $10^j$ as before

$$\begin{array}{c|c}j&1&2&3&4&5&6&7&8&9&10&11\\\hline\alpha&2.60&3.88&6.34&8.41&9.90&11.79&13.55&15.24&16.94&18.66&20.36\end{array}$$

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