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.
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 $$ \frac{T_n}{P_n} \sim \frac{n}{2 \log n} \sim \frac{n}{2 \log P_n} $$