I was observing

$T_n:=1,3,6,10,15,...$ and $P_n=2,3,5,7,...$ of these sequences.

$T_n={n(n+1)\over 2}$ is the triangular numbers and $P_n$ is the prime numbers

We came acrossed this relation between them as follow

$${T_n\over P_n}\sim {n\ln{\pi}\over 2\ln{P_n}}$$

Where $\pi=3.14159...$

The difference between them are small enough, so we assume this might happen as $n\to \infty$ $\lim_{n\to \infty}\left\lvert {T_n\over P_n}- {n\ln{\pi}\over 2\ln{P_n}} \right\rvert=1$

My question is: Does this relation ${T_n\over P_n}\sim {n\ln{\pi}\over 2\ln{P_n}}$ bring any usefulness to understanding the prime number problem?

Also can one show that ${T_n\over P_n}\sim {n\ln{\pi}\over 2\ln{P_n}}$ hold for $n \to \infty$ or it will jump off; such as the difference between them is so huge?.

This ${T_n\over P_n}\sim {n\ln{\pi}\over 2\ln{P_n}}$ can be simplify to

$$n+1\sim {P_n\ln{\pi}\over \ln{P_n}}$$ this means that given any known prime number we can approximate its nth location in the sequence.

  • $\begingroup$ I believe that there are different relationships between prime numbers and triangle numbers, but currently all my conclusions and attempts are useless. If $n$ is an odd prime number and $1+2+\ldots+ n=T_n$ is Gauss sum, then denoting the radical of an integer $m>1$ with $\operatorname{rad}(m)$ one can deduce (the useless statement) $$\operatorname{rad}(T_n)=n\cdot\operatorname{rad}\left(\frac{n+1}{2}\right).$$ I've tried think about this identity (and some topics that I believe that are related) but I concluded nothing useful. $\endgroup$ – user243301 Dec 23 '17 at 13:53
  • $\begingroup$ Thank you @user243301, no problem, you did your best, I am very grateful $\endgroup$ – user515439 Dec 23 '17 at 14:46

The $\log\pi$ factor seems extraneous.

Indeed, $ T_n \sim \dfrac{n^2}{2} $ and $ P_n \sim n \log n $ implies $$ \frac{T_n}{P_n} \sim \frac{n^2}{2 n \log n} = \frac{n}{2 \log n} $$ Now, $P_n \sim n \log n$ implies $\log P_n \sim \log n + \log \log n \sim \log n$ and so $$ \frac{T_n}{P_n} \sim \frac{n}{2 \log n} \sim \frac{n}{2 \log P_n} $$

  • $\begingroup$ I think OP was confused about the $\ln \pi$ factor since $\ln \pi \approx 1$, so it wouldn't change the right-hand-side of OP's asymptotic relation much. $\endgroup$ – Jam Dec 23 '17 at 12:40
  • 1
    $\begingroup$ @Jam, good point! $\endgroup$ – lhf Dec 23 '17 at 12:47

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