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Consider the sequence $u_n=n^a+p_n$, where $a$ is a real constant and $p_n$ is the $n$th prime.

Write down the terms in $u_n$ increasing from left to right. From each term, draw a line segment connecting it to its numerically closer neighbor, choosing the lower in case of a tie, thus forming clusters. Here is an example with $a=1$.

$\color{red}{3-5-8-11} \qquad \color{blue}{16-19} \qquad \color{brown}{24-27-32} \qquad \color{green}{39-42}\qquad \color{orange}{49-54-57-62-69} \qquad \color{purple}{76-79} \qquad \cdots$

The cluster sizes are $4,2,3,2,5,3,\dots$.

Is the following conjecture true:

Conjecture: The average cluster size is $3$ if $a\le 2$, and infinity if $a>2$.

My conjecture is based on numerical investigation with Excel, using the first $10^6$ primes. My conjecture can be expressed more precisely as follows. For a given value of $a$, let $f(a,c)$ be the arithmetic mean of the sizes of the first $c$ clusters. My conjecture is that, as $c\to\infty$, $f(a,c)$ approaches $3$ if $a\le 2$, and $f(a,c)$ approaches infinity if $a>2$.

Connection with Poisson distribution

In a $1D$ Poisson process, the probability that a point is its nearest neighbor's nearest neighbor, is $p=2/3$, as shown here, so the average cluster size is $2/p=3$, as shown here. Coming back to this question, I guess that $u_n$ resembles a $1D$ Poisson process in some critical way if $a\le 2$.

This question was inspired by the question linked to above.

Dan
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    By the prime number theorem, $p_n\sim n\log n$, so the gaps between primes are of the order of $\log n$. They also fluctuate on that order. The difference between successive differences of $n^a$ is of the order $n^{a-2}$. So for $a\gt2$ you're favouring downward links in a way that will eventually win out over any differences in the prime gaps, whereas for $a\ge2$ the bias is constant (for $a=2$) or goes to $0$ (for $a\lt2$), so it eventually gets drowned out by the fluctuations in the prime gaps. – joriki Jan 18 '24 at 13:25
  • @joriki Interesting! (I think you meant "... whereas for $a\le 2$..." – Dan Jan 18 '24 at 13:29
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    So the main content of your conjecture is that (heuristically speaking) the prime gaps are either "independent" (in which case they necessarily have the same cluster distribution as a Poisson process since in both cases the order of the gaps is a uniformly random permutation) or are dependent in precisely such a way that they reproduce the average cluster size (which seems unlikely). You could differentiate between these two possibilities by comparing the entire cluster size distribution with the one for the Poisson distribution. – joriki Jan 18 '24 at 13:31
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    Yes, sorry, for $a\le2$ – one minute too late to edit :-) – joriki Jan 18 '24 at 13:32
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    By the way, your other questions prompted me to look for the cluster size distribution of the Poisson process in one dimension. I was going to write that up as a self-answered question; I'll post a link here if/when I do that. – joriki Jan 18 '24 at 13:35
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    Also, note that unexpected biases in the primes can be quite subtle (see https://en.wikipedia.org/wiki/Chebyshev%27s_bias), so you might not pick them up with counts up to $10^6$. I'd suggest to drop the $n^a$ (which as far as I can tell is just a distraction from the interesting question) and directly test the equidistribution of order permutations among tuples of successive prime gaps. – joriki Jan 18 '24 at 14:07

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