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I have taken 2 sets: The first is a consecutive list of the first prime of twin pairs. The second is a consecutive list of numbers as follow 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5 ....

I have then compared between the lists by dividing the numbers of the second list with the numbers of the first list, and a steady growth rate of distribution occurs (as seen in the pictures below).

If you analyze the data (as seen in the pictures below), you will notice that:

If the fluctuation of column E is too high (usually above 1.1), then the "next" twin pair will have to be smaller than the "current:" pair, thus producing an error.

You can also notice that the fluctuation of column E is never too low (probably not less than 0.99 after the first few hundreds).

The same phenomenon happens if I replace Column C with the squares 1,4,9,16,… or with an arbitrary quadratic polynomial.

When replacing column C with a constant equal to 1, the max value never passes 1 (obviously). However, after the first few hundreds the min value again is probably not less than 0.99

Can anyone provide me with a theoretic explanation for why this might be?.

enter image description here enter image description here

List of first 100,000 with column C: 1, 1+2, 1+2+3, 1+2+3+4 ....

List of first 100,000 with column C: with the squares 1,4,9,16,25 ...

List of first 100,000 with column C: constant = 1

Thanks.

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    $\begingroup$ These numbers look way too small to predict any long-term behaviour. $\endgroup$
    – LSpice
    Aug 18, 2020 at 21:54
  • $\begingroup$ I have a list of the first 100,000, here is the link: numbersprime.com/crawlx.php $\endgroup$
    – Ilan Alon
    Aug 18, 2020 at 21:56
  • $\begingroup$ You do know there is a 'formula' (closed form) for the numbers in column C, right? $\endgroup$
    – Stopple
    Aug 18, 2020 at 22:53
  • $\begingroup$ Yes, thank you. I would like to clarify: my question is not whether column C grows infinitely or not. My question is what could possibly explain the minimum of 0.99 after the first 100? $\endgroup$
    – Ilan Alon
    Aug 18, 2020 at 23:12
  • $\begingroup$ You might experiment to see whether you get the same phenomenon if you replace Column C with the squares $1,4,9,16,\dots$ or with an arbitrary quadratic polynomial $\endgroup$ Aug 19, 2020 at 3:17

1 Answer 1

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What is the motivation of this tangle of computations?

Let $B_2=3,B_3=5,\cdots $ be your sequence of "first member of a twin prime pair." For some reason starting at index $2.$ We don't know that this is an infinite sequence but strongly suspect that it is with $B_n \approx k n (\ln n)^2$ for some constant $k.$ There are conjectures on $k$ but that hardly matters here. So for a plausible explanation we can say that $\frac{B_n}{B_{n-1}}$ is definitely greater than $1$ but approaching it at a steady average pace. Perhaps with $1<\frac{B_n}{B_{n-1}}<\frac{n+8}{n-1}.$ Or, to be especially reckless, $\frac{B_n}{B_{n-1}} \approx \frac{n}{n-1}.$

The numbers $E_n$ you are analyzing are exactly $\frac{B_n}{B_{n-1}}\frac{n-1}{n+1}$ so there is your explanation for why they are sometimes above $1$ and sometimes below, with convergence to $1.$


Digression: After the first few pairs, every member of the sequence is $11,17$ or $29 \bmod 30.$ Perhaps this introduces a little clumpiness. I don't know. You might check if the over vs under $1$ behavior correlates to congruence class $\bmod 30$ being $11$ vs $17$ or $29.$ If so, does this behavior seem to continue or die out?


The sequence $C_1=1,C_2=3,\cdots $ of triangular numbers has $C_n=\frac{n(n+1)}2$ so $\frac{C_{n-1}}{C_{n}}=\frac{n-1}{n+1}$ exactly.

You define $D_n=\frac{C_n}{B_n}$ and then, for $n \ge 3,$ $$E_n=\frac{D_n}{D_{n-1}}=\frac{B_n}{B_{n-1}}\frac{C_{n-1}}{C_n}=\frac{B_n}{B_{n-1}}\frac{n-1}{n+1}\approx\frac{n}{n-1}\frac{n-1}{n+1} \rightarrow 1$$

If instead of twin primes you used primes, with $p_n \approx n\ln n,$ results should be about the same, possibly less choppy. If instead of triangular numbers you used squares you would have $\frac{(n-1)^2}{n^2}\approx \frac{n-2}{n}$ which is very close to $\frac{n-1}{n+1}$

The further steps of adding successive terms of a previous column or taking ratios gives sequences which converge to one or grow like $n.$

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  • $\begingroup$ I understand your answer and accept your answer; yet I am sorry for my following naive question: I still don’t understand why the notion is that “Twin primes become increasingly scarce among larger numbers”, If the numbers in column E convergence to 1? $\endgroup$
    – Ilan Alon
    Aug 23, 2020 at 3:24
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    $\begingroup$ Squares are more predictable but MUCH rarer than twin primes. According to your data $B_{2000}=182,009$ $C_{2000}=2,003,001>10 B_{2000}.$ So for sure the squares get sparse. But their ratio goes to $1$ like $1+\frac2n.$ And the twin primes, while increasingly sparse (but not as sparse), have ratio going to $1$ like $1+\frac1n.$ Up to $x$ there are about $\sqrt{x}$ squares, $\frac{x}{\ln x}$ primes and $\frac{x}{\ln(x)^2}$ twin primes. $\endgroup$ Aug 23, 2020 at 3:58

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